Lesson Explainer: Curve Sketching using Derivatives Mathematics

In this explainer, we will learn how to use derivatives to graph different functions.

There are a lot of different techniques for sketching the graph of a function. For example, to sketch 𝑦=𝑓(π‘₯), we can solve 𝑓(π‘₯)=0 to find the π‘₯-intercepts; we know the 𝑦-intercept is 𝑓(0); we can try to find the horizontal and vertical asymptotes by considering limits of the function. However, this is not always enough information to construct an accurate sketch of our diagram.

One way of improving the accuracy of our sketch is to use some of the results we have from the first and second derivatives of the function. In other words, if we can find 𝑓′(π‘₯) and 𝑓′′(π‘₯), then we can determine a lot of information about our curve.

Properties: The First Derivative

If 𝑓′(π‘₯)=0 or does not exist, 𝑓 is said to have a critical point at π‘₯. In particular, if 𝑓′(π‘₯)=0, then there might be a turning point at this value of π‘₯. We can also use the first derivative test to determine the type of critical points we have; these could be local extrema, points of inflection, or points of discontinuity.

If 𝑓′(π‘₯)>0, where π‘₯∈]π‘Ž,𝑏[, then 𝑓 is increasing on the interval ]π‘Ž,𝑏[.

If 𝑓′(π‘₯)<0, where π‘₯∈]π‘Ž,𝑏[, then 𝑓 is decreasing on the interval ]π‘Ž,𝑏[.

If 𝑓 is continuous, then on any interval [π‘Ž,𝑏] the absolute extrema will occur at the critical points of 𝑓 or at the endpoints of the closed interval.

Similarly, if we can determine 𝑓′′(π‘₯), then we can determine even more information about our curve.

Properties: The Second Derivative

Consider a function 𝑦=𝑓(π‘₯), twice differentiable on ]π‘Ž,𝑏[.

If 𝑓′′(π‘₯)>0, for π‘₯∈]π‘Ž,𝑏[, then 𝑓 is concave upward on the interval ]π‘Ž,𝑏[.

If 𝑓′′(π‘₯)<0, for π‘₯∈]π‘Ž,𝑏[, then 𝑓 is concave downward on the interval ]π‘Ž,𝑏[.

If the function 𝑓(π‘₯) has a critical point when π‘₯=𝑐, then we can use the second derivative test to attempt to categorize this point.

  • If 𝑓′′(𝑐)>0, the point is a relative minimum.
  • If 𝑓′′(𝑐)<0, the point is a relative maximum.
  • If 𝑓′′(𝑐)=0, the point could be a relative minimum, a relative maximum, or a point of inflection. We will need to check the first derivative on either side of 𝑐 to determine its nature. To check if there is an inflection point at π‘₯=𝑐, we check if the sign of the second derivative switches on either side of 𝑐. If the sign does change, then 𝑓(π‘₯) has an inflection point at π‘₯=𝑐.

When we are asked to sketch 𝑦=𝑓(π‘₯), we can also check to see where the function is differentiable and what its derivatives are to help us sketch the curve. Let us start by listing all of the properties we can check to reveal information about the function and its graph.

How To: Graphing Using Derivatives

For a curve 𝑦=𝑓(π‘₯), if possible, we can check the following properties:

  1. Domain and Range
    The domain of a function is the set of values for which the function is defined. Graphically, this tells us the set of π‘₯-values over which our curve will be sketched.
    The range of a function is the set of all possible outputs of our function given the domain. Graphically, this will tell us the set of possible 𝑦-values for our curve.
  2. Intercepts
    We can find the π‘₯-intercepts of our curve by solving the equation 𝑓(π‘₯)=0.
    If 0 is in the domain of our function, then we can find the 𝑦-intercept by evaluating 𝑓(0).
  3. Symmetry
    If 𝑓(βˆ’π‘₯)=𝑓(π‘₯) for all values of π‘₯ in the domain of our function, then we say the function is even. Even functions have reflectional symmetry about the 𝑦-axis.
    If 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯) for all values of π‘₯ in the domain our function, then we say the function is odd. All odd functions map onto themselves when the curve is rotated 180∘ about the origin.
  4. Intervals of Increase or Decrease
    If 𝑓′(π‘₯)>0 on an interval, then 𝑓 is increasing on that interval.
    If 𝑓′(π‘₯)<0 on an interval, then 𝑓 is decreasing on that interval.
  5. Local Extrema/Turning Points
    We know that local extrema and turning points must occur at the critical points of our function (where 𝑓′(π‘₯)=0, does not exist, or is for nonopen domains at the endpoints).
    We can use this to find the π‘₯-values of possible local extrema and then use either the first derivative test, the second derivative test, or inspection to decide on their nature.
  6. Concavity and Points of Inflection
    We can determine the concavity of our curve if it is twice differentiable on an interval.
    If 𝑓′′(π‘₯)>0 on an interval, then the curve 𝑦=𝑓(π‘₯) is concave upward on that interval.
    If 𝑓′′(π‘₯)<0 on an interval, then the curve 𝑦=𝑓(π‘₯) is concave downward on that interval.
    We should also be aware that points of inflection occur where the concavity changes. In other words, the function changes from concave upward to downward, or vice versa.
  7. Asymptotes/End Behavior
    Finally, we can look for asymptotes and at the end behavior of our curve.
    Remember, we have a vertical asymptote at π‘Ž if
    • limο—β†’οŒΊοŽͺ𝑓(π‘₯)=±∞,
    • limο—β†’οŒΊοŽ©π‘“(π‘₯)=±∞,
    • limο—β†’οŒΊπ‘“(π‘₯)=±∞.

There are a few options for what can happen to our curve as π‘₯ approaches positive or negative infinity, which is called the end behavior of our curve. The best way to determine this is by considering limο—β†’βˆžπ‘“(π‘₯) and limο—β†’οŠ±βˆžπ‘“(π‘₯).

One option is to have a horizontal asymptote; a horizontal asymptote 𝑦=𝐿 occurs when limο—β†’βˆžπ‘“(π‘₯)=𝐿 or limο—β†’οŠ±βˆžπ‘“(π‘₯)=𝐿.

Another option is one (or both) of these limits approaching positive or negative infinity. Of course, these are not the only options for the end behavior of our curve, but by looking at these limits, we can usually deduce the end behavior.

We will not need to consider every element in this list for every function we are asked to sketch; we only need to check the properties that will help us sketch the curve.

In our first two examples, we will look at how to apply these guidelines to polynomial functions.

Example 1: Identifying the Correct Graph of a Cubic Polynomial

Use derivatives to identify which of the following is the graph of the function 𝑓(π‘₯)=βˆ’π‘₯+6π‘₯βˆ’9π‘₯+1.

Answer

In this question, we are asked to identify the correct sketch of 𝑓(π‘₯)=βˆ’π‘₯+6π‘₯βˆ’9π‘₯+1 by using derivatives. To do this, we will start with everything we can deduce about the curve without derivatives.

First, all polynomials are defined for any input value of π‘₯, so the domain of 𝑓(π‘₯) is all real values of π‘₯. Since this function is a polynomial, we can get an idea of the range of 𝑓 by looking at the leading term. We know that the shape of our curve will be a cubic with negative leading coefficient. In particular, since this polynomial has an odd degree, its range is all real numbers.

Second, we can find the 𝑦-intercept by substituting π‘₯=0 into the original function, 𝑓(0)=βˆ’0+6(0)βˆ’9(0)+1=1.

So, our 𝑦-intercept is 1.

We could try to find the π‘₯-intercepts by solving 𝑓(π‘₯)=0; however, we cannot find any roots by inspection. We know there must be at least 1 root since the range of 𝑓 is all real numbers. If we were tasked with sketching this graph ourselves, we could use our calculator to find the exact value of the roots. However, we only need to identify the graph, so this is not necessary.

Next, we can find more information about our curve by using derivatives.

We can find the coordinates of the turning points to better give us an idea of the shape of the graph. To do this, we want to solve 𝑓′(π‘₯)=0: 𝑓′(π‘₯)=βˆ’3π‘₯+12π‘₯βˆ’9βˆ’3π‘₯+12π‘₯βˆ’9=0βˆ’3(π‘₯βˆ’3)(π‘₯βˆ’1)=0.

We can see we have two critical points, one when π‘₯=1 and the other when π‘₯=3. Since this is a polynomial, we know its derivative exists for all real values, so these will be our only critical points.

We can check if these are local extrema by evaluating the slope of our curve around these values:

π‘₯01234
𝑓′(π‘₯)βˆ’9030βˆ’9

Since the slope changes from negative to positive about π‘₯=1, there must be a local minimum here. Similarly, the slope changes from positive to negative about π‘₯=3, so this point is a local maximum. Substituting these values into the original function, we can find the coordinates of these two points: 𝑓(1)=βˆ’3,𝑓(3)=1.

So far, we have found all of the turning points and the 𝑦-intercept of our curve, and we have also shown it will have the shape of a cubic polynomial with a negative leading coefficient.

We can also find the second derivative of the curve to determine its convexity. We use the power rule for differentiation to find the second derivative as 𝑓′′(π‘₯)=βˆ’6π‘₯+12.

We can then determine the points where the convexity changes by considering the sign of the second derivative. We see that βˆ’6π‘₯+12=0π‘₯=2, so the convexity of the curve changes around π‘₯=2. In particular, the second derivative is positive when π‘₯<2 and negative when π‘₯>2, so the curve is convex upward when π‘₯>2 and convex downward when π‘₯<2. Since the convexity changes at π‘₯=2, we can say that this is an inflection point of the curve.

This is enough to sketch our curve.

We can see this is given as option D.

Let’s see another example of how we can apply this process to more accurately sketch the graphs of a polynomial curve.

Example 2: Sketching the Graph of a Polynomial Using Derivatives

Consider the function 𝑓(π‘₯)=(π‘₯βˆ’1)(π‘₯+2).

  1. Find 𝑓′(π‘₯).
  2. Find and classify the critical points of 𝑓.
  3. Find the intervals of increase and decrease for 𝑓.
  4. Find limο—β†’βˆžπ‘“(π‘₯).
  5. Which of the following is the graph of 𝑓?

Answer

This question walks us through a process to fully sketch the graph of the polynomial function 𝑓(π‘₯)=(π‘₯βˆ’1)(π‘₯+2).

Part 1

We first want to find the derivative of this function and there are a few ways of doing this. We will use the product rule and general power rule: 𝑓′(π‘₯)=[2(π‘₯βˆ’1)](π‘₯+2)+(π‘₯βˆ’1)[1]=2(π‘₯βˆ’1)(π‘₯+2)+(π‘₯βˆ’1)=(π‘₯βˆ’1)(2π‘₯+4+π‘₯βˆ’1)=(π‘₯βˆ’1)(3π‘₯+3)=3(π‘₯βˆ’1)(π‘₯+1).

Distributing the parentheses, we get 𝑓′(π‘₯)=3π‘₯βˆ’3.

It is worth noting that since 𝑓(π‘₯) is a polynomial, all the derivatives of 𝑓(π‘₯) will be polynomials.

Part 2

Next, the question wants us to find and classify the critical points of our function 𝑓.

Remember, critical points are when the derivative is zero or does not exist; in our case, 𝑓 is a polynomial, so its derivative always exists. This means the only critical points occur when the derivative is equal to zero.

We will set the factored expression for 𝑓′(π‘₯) equal to zero and solve: 3(π‘₯βˆ’1)(π‘₯+1)=0.

Thus, we must have two critical points: π‘₯=1 and π‘₯=βˆ’1.

We can classify these critical points by using the first derivative test. We evaluate the slope of the function on either side of the critical point to determine its nature.

π‘₯βˆ’2βˆ’1012
𝑓′(π‘₯)90βˆ’309

A change from a positive to a negative slope means that we have a local maximum at π‘₯=βˆ’1, and a change from a negative to a positive slope means we have a local minimum at π‘₯=1.

We can then evaluate the function at these points to find the coordinates of its local extrema: 𝑓(1)=0,𝑓(βˆ’1)=4.

Therefore, 𝑓 has a local minimum at (1,0) and a local maximum at (βˆ’1,4).

Part 3

Next, we are tasked with finding the intervals of increase and decrease for our function. To do this, we recall that, for a differentiable function, we know if 𝑓′(π‘₯)>0 on an interval, then the function is increasing on that interval. Similarly, if 𝑓′(π‘₯)<0 on an interval, then 𝑓 is decreasing on that interval. We know 𝑓 is differentiable for all real values since it is a polynomial, so we need to find the intervals where 𝑓′(π‘₯) is positive and negative.

There are a few ways of doing this. We know 𝑓′(π‘₯)=3(π‘₯βˆ’1)(π‘₯+1) has a positive leading coefficient quadratic with π‘₯-interceps at 1 and βˆ’1, so we can sketch this function.

We see that 𝑓′(π‘₯) is positive when π‘₯∈]βˆ’βˆž,βˆ’1[ and when π‘₯∈]1,∞[. We can also see 𝑓′(π‘₯) is negative when π‘₯∈]βˆ’1,1[. These tell us the sign of the slope of 𝑓 on these intervals and, hence, the intervals where 𝑓 is increasing or decreasing.

Therefore, 𝑓 is increasing on ]βˆ’βˆž,βˆ’1[ and ]1,∞[ and decreasing on ]βˆ’1,1[.

Part 4

The next part of this question asks us to evaluate limο—β†’βˆžπ‘“(π‘₯); to do this, we need to evaluate its limit as π‘₯β†’βˆž and π‘₯β†’βˆ’βˆž; we will start with the former: limlimlimο—β†’βˆžο—β†’βˆžοŠ¨ο—β†’βˆžοŠ©π‘“(π‘₯)=(π‘₯βˆ’1)(π‘₯+2)=π‘₯βˆ’3π‘₯+2.

Since 𝑓 is a polynomial, its end behavior is decided by its leading term: limlimο—β†’βˆžοŠ©ο—β†’βˆžοŠ©π‘₯βˆ’3π‘₯+2=π‘₯.

We can then evaluate this limit directly. As π‘₯ approaches infinity, π‘₯ is growing without bound, which means limο—β†’βˆžπ‘“(π‘₯)=∞.

It is also worth pointing out that we can do the same as π‘₯ approaches βˆ’βˆž: limlimlimlimο—β†’οŠ±βˆžο—β†’οŠ±βˆžοŠ¨ο—β†’βˆžοŠ©ο—β†’οŠ±βˆžοŠ©π‘“(π‘₯)=(π‘₯βˆ’1)(π‘₯+2)=π‘₯βˆ’3π‘₯+2=π‘₯=βˆ’βˆž.

Part 5

The final part of this question asks us to identify the correct sketch of the curve 𝑦=𝑓(π‘₯). We could do this by elimination; however, we almost have enough information to completely sketch our curve.

We can find the π‘₯-intercepts by solving 𝑓(π‘₯)=0. We know that 𝑓(π‘₯)=(π‘₯βˆ’1)(π‘₯+2), so the π‘₯-intercepts occur when (π‘₯βˆ’1)(π‘₯+2)=0.

These are π‘₯=1 and π‘₯=βˆ’2. We can also find the 𝑦-intercept at 𝑓(0)=2.

We also know there is a local minimum at (1,0) and a local maximum at (βˆ’1,4).

We can determine the convexity of the curve by considering the sign of its second derivative. We can differentiate the expression we found in part 1 by using the power rule for differentiation to get 𝑓′′(π‘₯)=6π‘₯.

We then see that 𝑓′′(π‘₯) is positive when π‘₯>0, so it is convex downward on this interval, and that 𝑓′′(π‘₯) is negative when π‘₯<0, so it is convex upward on this interval.

Finally, we know limο—β†’βˆžπ‘“(π‘₯)=∞ and limο—β†’οŠ±βˆžπ‘“(π‘₯)=βˆ’βˆž.

This gives us the following sketch.

We can see the correct curve is option D.

So far, we have only seen examples involving polynomial functions. However, our methods can work on a variety of functions. Let’s see some examples involving rational functions.

Example 3: Graphing a Rational Function Using Derivatives

Consider the function 𝑓(π‘₯)=4π‘₯π‘₯+3.

  1. Find all asymptotes of 𝑓.
  2. Find 𝑓′(π‘₯).
  3. Find and classify all critical points of 𝑓.
  4. Find the intervals of increase and decrease for 𝑓.
  5. Which of the following could be the graph of 𝑓?

Answer

Part 1

The parts of this question will give us the tools we need to sketch the graph 𝑦=𝑓(π‘₯).

The first part asks us to find all of the asymptotes of the rational function 𝑓. There are a few options for doing this; we could directly use the definitions of asymptotes; however, since 𝑓 is a rational function, we can find the asymptotes using a few rules.

Let 𝑓(π‘₯)=𝑃(π‘₯)𝑄(π‘₯) be a rational function with polynomials 𝑃 and 𝑄. To find the vertical asymptotes of 𝑓, we first need to find the values of π‘₯ for which the denominator 𝑄(π‘₯) is zero.

In our function, 𝑃(π‘₯)=4π‘₯ and 𝑄(π‘₯)=π‘₯+3. We will start by solving 𝑄(π‘₯)=0.

The discriminant of this quadratic is 0βˆ’4(1)(3)=βˆ’12; since this is negative, there are no real roots to this quadratic. Hence, 𝑓(π‘₯) has no vertical asymptotes.

It is worth noting that since the denominator of 𝑓(π‘₯) is never 0, 𝑓 is defined for all real values of π‘₯. In other words, its domain is ℝ.

To find the horizontal asymptotes of a rational function, we need to compare the degrees of 𝑃 and 𝑄. We can see 𝑃 and 𝑄 are quadratic functions, which means their degrees are both 2. So, the numerator and denominator have the same degree; in this case, we will have a horizontal asymptote. We find this by evaluating the following limit: limlimο—β†’βˆžο—β†’βˆžοŠ¨οŠ¨π‘“(π‘₯)=4π‘₯π‘₯+3.

We evaluate this by rewriting 𝑓(π‘₯) by using algebraic division or inspection: limlimlimlimο—β†’βˆžοŠ¨οŠ¨ο—β†’βˆžοŠ¨οŠ¨ο—β†’βˆžοŠ¨οŠ¨οŠ¨ο—β†’βˆžοŠ¨4π‘₯π‘₯+3=ο€Ύ4π‘₯+12βˆ’12π‘₯+3=ο€Ύ4π‘₯+12π‘₯+3βˆ’12π‘₯+3=ο€Ό4βˆ’12π‘₯+3=4.

We can do the same for π‘₯β†’βˆ’βˆž; however, we will get the same result: limlimο—β†’οŠ±βˆžο—β†’οŠ±βˆžοŠ¨π‘“(π‘₯)=ο€Ό4βˆ’12π‘₯+3=4.

Hence, the only asymptote of 𝑦=𝑓(π‘₯) is the line 𝑦=4, which is a horizontal asymptote on both sides.

Part 2

Next, we are asked to find an expression for 𝑓′(π‘₯). There are few options for doing this; we will use the quotient rule: 𝑓′(π‘₯)=ο€Ήπ‘₯+3⋅(8π‘₯)βˆ’ο€Ή4π‘₯⋅(2π‘₯)(π‘₯+3)=8π‘₯+24π‘₯βˆ’8π‘₯(π‘₯+3)=24π‘₯(π‘₯+3).

This means 𝑓′(π‘₯)=24π‘₯(π‘₯+3).

Part 3

The third part of this question wants us to find and classify all of the critical points of 𝑓(π‘₯).

First, we need to remember that the critical points of 𝑓 are the values of π‘₯ in the domain of 𝑓 where 𝑓′(π‘₯)=0 or does not exist.

So, to find the critical points of 𝑓, we can start by solving 𝑓′(π‘₯)=0, this means the numerator of the expression we found for 𝑓′ must be zero: 24π‘₯=0,π‘₯=0.whichgives

Therefore, 𝑓 has a critical point at π‘₯=0.

Next, we can see if there are any points where the derivative is not defined, the only way this could happen is if the denominator of 𝑓′ is zero. This means ο€Ήπ‘₯+3=0.

We have already shown this quadratic has no real roots, so the only critical point is when π‘₯=0.

We can determine the nature of these critical points by using the first derivative test.

π‘₯βˆ’101
𝑓′(π‘₯)βˆ’1.501.5

We can see the slope of 𝑓 changes from negative to positive about π‘₯=0, which means 𝑓 must have a local minimum at this point.

Part 4

The fourth part of this question wants us to determine the intervals of increase and decrease for our function. Since our function is differentiable across its entire domain, we can do this by checking where the slope is positive and where it is negative.

To do this, lets first look at 𝑓′(π‘₯): 𝑓′(π‘₯)=24π‘₯(π‘₯+3).

We can see the denominator is a squared expression; this means it will never be negative, so the sign of 𝑓′ is decided entirely by the numerator. Hence, we can find where 𝑓′(π‘₯)<0 by solving the inequality 24π‘₯<0π‘₯<0.

Similarly, we can find where 𝑓′(π‘₯)>0 by solving the inequality 24π‘₯>0π‘₯>0.

This means that 𝑓 is increasing on (0,∞) and decreasing on (βˆ’βˆž,0).

Part 5

The final part of this question wants us to choose the correct sketch of 𝑓(π‘₯); however, we can also try to sketch this curve ourselves.

First, we can add in the asymptote of our graph.

Next, we can find the 𝑦-intercept of our curve: 𝑓(0)=0, so the curve has a 𝑦-intercept of 0.

We can find the π‘₯-intercept: 𝑓(π‘₯)=04π‘₯π‘₯+3=04π‘₯=0π‘₯=0.

Hence, the curve has its only π‘₯-intercept at 0.

We can also find the coordinates of the local minima by substituting the π‘₯-value of the critical points into the original function: 𝑓(0)=0.

So, the point (0,0) is a local minimum and is the only turning point of the curve.

We can also find the second derivative of the curve to determine its convexity. We will differentiate the expression we found in part 2 by using the quotient rule. This gives us 𝑓′′(π‘₯)=ο€Ήπ‘₯+3⋅24βˆ’2(2π‘₯)ο€Ήπ‘₯+324π‘₯ο€Ί(π‘₯+3)=24ο€Ήπ‘₯+3ο…βˆ’96π‘₯ο€Ήπ‘₯+3(π‘₯+3)=24ο€Ήπ‘₯+3ο…βˆ’96π‘₯ο€Ήπ‘₯+3(π‘₯+3).οŠͺ

We then cancel the shared factor of ο€Ήπ‘₯+3ο…οŠ¨ and simplify, where we note that π‘₯+3>0, so it is never equal to zero. This gives 𝑓′′(π‘₯)=24ο€Ήπ‘₯+3ο…βˆ’96π‘₯(π‘₯+3)=24π‘₯+72βˆ’96π‘₯(π‘₯+3)=βˆ’72π‘₯+72(π‘₯+3)=βˆ’72ο€Ήπ‘₯βˆ’1(π‘₯+3).

We can then determine the points where the convexity changes by considering the sign of the second derivative. We see that π‘₯+3 is always positive, so the denominator does not affect the sign of the second derivative. Hence, we only need to consider the sign of the numerator.

We note that π‘₯βˆ’1 is negative when βˆ’1<π‘₯<1 and it is positive when π‘₯<βˆ’1 or π‘₯>1. Thus, the second derivative of the curve is positive when βˆ’1<π‘₯<1, and so it is convex downward on this interval, and the second derivative of the curve is negative when π‘₯<βˆ’1 or π‘₯>1, and so it is convex upward on these intervals. We also note that the convexity of the curve changes at π‘₯=βˆ’1 and π‘₯=1, so these are points of inflection.

This gives us the following sketch.

The correct graph is option D.

Example 4: Identifying the Correct Graph of a Rational Function

Use derivatives to identify which of the following is the graph of the function 𝑓(π‘₯)=110π‘₯+10π‘₯.

Answer

Since the question wants us to sketch this graph by using derivatives, we should start by collecting as much information as we need to sketch the curve.

First, since this is a rational function, the domain is all values of π‘₯ except those that make the denominator equal to 0. This means we can find the values of π‘₯ not in the domain of our function by making the denominator equal to zero and solving: 10π‘₯+10π‘₯=010π‘₯(π‘₯+1)=0.

So, the domain of 𝑓 is all real numbers except 0 and βˆ’1.

Next, we can find the intercepts. To find the 𝑦-intercept, we substitute π‘₯=0 into 𝑓𝑓(0)=110(0)+10(0)=10.

Therefore, 0 is not in the domain of 𝑓 and 𝑓 does not have a 𝑦-intercept. In fact, this shows 𝑓 has a vertical asymptote at 0.

To find the π‘₯-intercepts of our curve, we need to find the values of π‘₯ that satisfy the equation 𝑓(π‘₯)=0.

For this to happen, the numerator of our rational function would need to be 0, but this can never happen, since the numerator is always equal to 1. Therefore, 𝑓 has no π‘₯-intercepts either.

We can then find the local extrema. We will do this by finding the critical points. First, let us find 𝑓′(π‘₯) by using the quotient rule: 𝑓′(π‘₯)=βˆ’20π‘₯βˆ’10(10π‘₯+10π‘₯).

The critical points will occur when 𝑓′(π‘₯)=0 or does not exist.

We can see 𝑓′(π‘₯)=0 when the numerator is equal to 0, which is when π‘₯=βˆ’12. We can see the only time 𝑓′(π‘₯) does not exist is when the denominator is equal to 0; however, this would also mean that 𝑓(π‘₯) does not exist, so this value of π‘₯ is not in the domain of 𝑓. This means our only critical point is π‘₯=βˆ’12.

We can find the type of critical point by using the first derivative test, that is, considering the sign of the slope of our curve on either side of the critical point.

It is important to remember π‘₯=βˆ’1 and π‘₯=0 are not in the domain of 𝑓, so we should not choose our sample points beyond these values.

π‘₯βˆ’34βˆ’12βˆ’14
𝑓′(π‘₯)Positive0Negative

We can see that the slope of 𝑓 changes from positive to negative about π‘₯=βˆ’12, so 𝑓 must have a local maximum at this point.

We can find the coordinates of this local maximum by evaluating the original function at π‘₯=βˆ’12; this gives us π‘“ο€Όβˆ’12=βˆ’25.

We can also find the intervals of increase and decrease for 𝑓 by checking where 𝑓′(π‘₯) is positive and negative.

To do this, let’s start with 𝑓′(π‘₯): 𝑓′(π‘₯)=βˆ’20π‘₯βˆ’10(10π‘₯+10π‘₯).

The denominator of this function is a square so it is never negative, for any real value of π‘₯; hence, the sign of 𝑓′(π‘₯) is entirely determined by the numerator.

We can then find these values of π‘₯: βˆ’20π‘₯βˆ’10>0βˆ’20π‘₯>10π‘₯<βˆ’12.

Therefore, for values of π‘₯ in the domain of 𝑓, 𝑓′(π‘₯) is positive when π‘₯<βˆ’12 and negative when π‘₯>βˆ’12.

So, 𝑓 is increasing when π‘₯<βˆ’12 and decreasing when π‘₯>βˆ’12 (and π‘₯ is in the domain of 𝑓).

We can determine the convexity of the curve by considering the sign of its second derivative. We can differentiate the first derivative by using the quotient rule to get 𝑓′′(π‘₯)=π‘₯(π‘₯+1)ο€Ή3π‘₯+3π‘₯+15π‘₯(π‘₯+1).οŠͺοŠͺ

We have not simplified the expression fully, so we can note that the denominator is a fourth power, and so it is positive on the domain of 𝑓. This means the sign of the second derivative is determined by the numerator of the expression. We then note that there are no real roots of the quadratic 3π‘₯+3π‘₯+1 since its discriminant is negative. In particular, we can use this to conclude the quadratic is always positive. Thus, the sign of the second derivative is determined by π‘₯(π‘₯+1).

We note that this is positive if π‘₯>0 or π‘₯<βˆ’1 and negative if βˆ’1<π‘₯<0. Hence, the curve is convex downward when π‘₯>0 or π‘₯<βˆ’1 and convex upward when βˆ’1<π‘₯<0. We can also note that the convexity of the curve changes at π‘₯=βˆ’1 and π‘₯=0, so these are points of inflection.

We can also find the asymptotes of 𝑓(π‘₯). We know rational functions have vertical asymptotes when the denominator is zero, but we should also check the numerator is nonzero. We already found these values when checking the domain. There are two vertical asymptotes, π‘₯=0 and π‘₯=βˆ’1.

The degree of the numerator is higher than the degree of the denominator, so as π‘₯ approaches positive or negative infinity, 𝑓(π‘₯) will approach 0, so 𝑦=0 is the horizontal asymptote of 𝑓(π‘₯) on both sides.

This is enough to sketch our curve.

Remember, the curve does not intersect the axes, the turning point is a local maximum, and the curve is increasing when π‘₯<βˆ’12 and decreasing when π‘₯>βˆ’12.

We can see this sketch is given as option A.

Let us finish by reviewing some of the key points when sketching curves of functions.

Key Points

  • There are a lot of properties we can check to help us sketch the curves of various functions.
  • We do not need to check every property, only the ones that will help us sketch the function.
  • It is almost always a good idea to look for the domain, intercepts, critical points, local extrema, asymptotes, and end behavior.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.