Lesson Explainer: Curve Sketching using Derivatives Mathematics

Example 1: Identifying the Correct Graph of a Cubic Polynomial

Use derivatives to identify which of the following is the graph of the function .

Answer

In this question, we are asked to identify the correct sketch of 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 , so the domain of is all real values of . Since this function is a polynomial, we can get an idea of the range of 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 -intercept by substituting into the original function,

So, our -intercept is 1.

We could try to find the -intercepts by solving ; however, we cannot find any roots by inspection. We know there must be at least 1 root since the range of 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 :

We can see we have two critical points, one when and the other when . 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:

Since the slope changes from negative to positive about , there must be a local minimum here. Similarly, the slope changes from positive to negative about , so this point is a local maximum. Substituting these values into the original function, we can find the coordinates of these two points:

So far, we have found all of the turning points and the -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

We can then determine the points where the convexity changes by considering the sign of the second derivative. We see that so the convexity of the curve changes around . In particular, the second derivative is positive when and negative when , so the curve is convex upward when and convex downward when . Since the convexity changes at , 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.

Example 2: Sketching the Graph of a Polynomial Using Derivatives

Consider the function .

1. Find .
2. Find and classify the critical points of .
3. Find the intervals of increase and decrease for .
4. Find .
5. Which of the following is the graph of ?

Answer

This question walks us through a process to fully sketch the graph of the polynomial function .

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:

Distributing the parentheses, we get

It is worth noting that since is a polynomial, all the derivatives of will be polynomials.

Part 2

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

Remember, critical points are when the derivative is zero or does not exist; in our case, 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 equal to zero and solve:

Thus, we must have two critical points: and .

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.

A change from a positive to a negative slope means that we have a local maximum at , and a change from a negative to a positive slope means we have a local minimum at .

We can then evaluate the function at these points to find the coordinates of its local extrema:

Therefore, has a local minimum at and a local maximum at .

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 on an interval, then the function is increasing on that interval. Similarly, if on an interval, then is decreasing on that interval. We know is differentiable for all real values since it is a polynomial, so we need to find the intervals where is positive and negative.

There are a few ways of doing this. We know has a positive leading coefficient quadratic with -interceps at 1 and , so we can sketch this function.

We see that is positive when and when . We can also see is negative when . These tell us the sign of the slope of on these intervals and, hence, the intervals where is increasing or decreasing.

Therefore, is increasing on and and decreasing on .

Part 4

The next part of this question asks us to evaluate ; to do this, we need to evaluate its limit as and ; we will start with the former:

Since is a polynomial, its end behavior is decided by its leading term:

We can then evaluate this limit directly. As approaches infinity, is growing without bound, which means

It is also worth pointing out that we can do the same as approaches :

Part 5

The final part of this question asks us to identify the correct sketch of the curve . We could do this by elimination; however, we almost have enough information to completely sketch our curve.

We can find the -intercepts by solving . We know that so the -intercepts occur when

These are and . We can also find the -intercept at .

We also know there is a local minimum at and a local maximum at .

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

We then see that is positive when , so it is convex downward on this interval, and that is negative when , so it is convex upward on this interval.

Finally, we know and .

This gives us the following sketch.

We can see the correct curve is option D.

Example 3: Graphing a Rational Function Using Derivatives

Consider the function .

1. Find all asymptotes of .
2. Find .
3. Find and classify all critical points of .
4. Find the intervals of increase and decrease for .
5. Which of the following could be the graph of ?

Answer

Part 1

The parts of this question will give us the tools we need to sketch the graph .

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

Let be a rational function with polynomials and . To find the vertical asymptotes of , we first need to find the values of for which the denominator is zero.

In our function, and . We will start by solving .

The discriminant of this quadratic is ; since this is negative, there are no real roots to this quadratic. Hence, has no vertical asymptotes.

It is worth noting that since the denominator of is never 0, is defined for all real values of . In other words, its domain is .

To find the horizontal asymptotes of a rational function, we need to compare the degrees of and . We can see and 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:

We evaluate this by rewriting by using algebraic division or inspection:

We can do the same for ; however, we will get the same result:

Hence, the only asymptote of is the line , which is a horizontal asymptote on both sides.

Part 2

Next, we are asked to find an expression for . There are few options for doing this; we will use the quotient rule:

This means

Part 3

The third part of this question wants us to find and classify all of the critical points of .

First, we need to remember that the critical points of are the values of in the domain of where or does not exist.

So, to find the critical points of , we can start by solving , this means the numerator of the expression we found for must be zero:

Therefore, has a critical point at .

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 is zero. This means

We have already shown this quadratic has no real roots, so the only critical point is when .

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

We can see the slope of changes from negative to positive about , which means 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 :

We can see the denominator is a squared expression; this means it will never be negative, so the sign of is decided entirely by the numerator. Hence, we can find where by solving the inequality

Similarly, we can find where by solving the inequality

This means that is increasing on and decreasing on .

Part 5

The final part of this question wants us to choose the correct sketch of ; 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 -intercept of our curve: so the curve has a -intercept of 0.

We can find the -intercept:

Hence, the curve has its only -intercept at 0.

We can also find the coordinates of the local minima by substituting the -value of the critical points into the original function:

So, the point 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

We then cancel the shared factor of and simplify, where we note that , so it is never equal to zero. This gives

We can then determine the points where the convexity changes by considering the sign of the second derivative. We see that 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 is negative when and it is positive when or . Thus, the second derivative of the curve is positive when , and so it is convex downward on this interval, and the second derivative of the curve is negative when or , and so it is convex upward on these intervals. We also note that the convexity of the curve changes at and , 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 .

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 except those that make the denominator equal to 0. This means we can find the values of not in the domain of our function by making the denominator equal to zero and solving:

So, the domain of is all real numbers except 0 and .

Next, we can find the intercepts. To find the -intercept, we substitute into

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

To find the -intercepts of our curve, we need to find the values of that satisfy the equation .

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, has no -intercepts either.

We can then find the local extrema. We will do this by finding the critical points. First, let us find by using the quotient rule:

The critical points will occur when or does not exist.

We can see when the numerator is equal to 0, which is when . We can see the only time does not exist is when the denominator is equal to 0; however, this would also mean that does not exist, so this value of is not in the domain of . This means our only critical point is .

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 and are not in the domain of , so we should not choose our sample points beyond these values.

We can see that the slope of changes from positive to negative about , so must have a local maximum at this point.

We can find the coordinates of this local maximum by evaluating the original function at ; this gives us

We can also find the intervals of increase and decrease for by checking where is positive and negative.

To do this, let’s start with :

The denominator of this function is a square so it is never negative, for any real value of ; hence, the sign of is entirely determined by the numerator.

We can then find these values of :

Therefore, for values of in the domain of , is positive when and negative when .

So, is increasing when and decreasing when (and is in the domain of ).

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

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 . 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 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 .

We note that this is positive if or and negative if . Hence, the curve is convex downward when or and convex upward when . We can also note that the convexity of the curve changes at and , so these are points of inflection.

We can also find the asymptotes of . 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, and .

The degree of the numerator is higher than the degree of the denominator, so as approaches positive or negative infinity, will approach 0, so is the horizontal asymptote of 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 and decreasing when .

We can see this sketch is given as option A.