Video: MATH-DIFF-INT-2018-S1-Q18A

What are the local maximum and minimum values of the function 𝑓(π‘₯) = π‘₯Β³ βˆ’ 3π‘₯ βˆ’ 2? Find the inflection point of the curve of the function, if it exists.

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Video Transcript

What are the local maximum and minimum values of the function 𝑓 of π‘₯ is equal to π‘₯ cubed minus three π‘₯ minus two? Find the inflection point of the curve of the function, if it exists.

In order to find the local maximum and minimum points of a function, we first need to find the critical points and then we decide whether these critical points are maximum or minimum. In order to find the critical point, we need to differentiate our function 𝑓 of π‘₯ and set this equal to zero. The differential of our function which weβ€²ll call 𝑓 dash of π‘₯ is also equal to d by dπ‘₯ of π‘₯ cubed minus three π‘₯ minus two.

Now, π‘₯ cubed minus three π‘₯ minus two is a polynomial. And in order to differentiate a polynomial, we simply take each term, multiply the term by the power, and then decrease the power by one. Our first term is π‘₯ cubed. And so, we multiply by the power to give us three. We then reduce the power of π‘₯ by one to give us π‘₯ squared. And so, π‘₯ cubed differentiates to three π‘₯ squared. For the second term, we have negative three π‘₯. The power of π‘₯ here is simply one. And so, we multiply by this power which is one. And we are left with the constant from before, which is negative three. We then reduce the power of π‘₯ by one to get π‘₯ to the power of zero. However, π‘₯ to the power of zero is equal to one. And so here, we get negative three multiplied by one which is simply negative three. The last term here negative two is a constant. And when we differentiate a constant, we simply get zero. And adding zero here would do nothing. So we can leave it out. Therefore, we found that 𝑓 dash of π‘₯ is equal to three π‘₯ squared minus three.

Now remember that we can find our critical point by setting 𝑓 dash of π‘₯ equal to zero. And so, therefore, we have that three π‘₯ squared minus three is equal to zero. We notice that each term here is a multiple of three. So we divide the whole equation by three. This gives us π‘₯ squared minus one is equal to zero. Adding ones to both sides of the equation gives us π‘₯ squared is equal to one. And now, we simply take square roots of both sides of the equation, giving us that the critical points are at π‘₯ is equal to plus or minus one. Now, since we took a square root here, we mustnβ€²t forget that we have both positive and negative roots, which is what’s gives us the plus or minus here.

Now that we have the critical points of our function, we need to find out whether they’re local maximum or minimum. There are a few ways in which we can do this. One way in which we can do this is to take the second differential of 𝑓, which we’ll call 𝑓 double dash of π‘₯. And then, if the value of 𝑓 double dash of our critical point is positive, then that point is a minimum. And if the 𝑓 double dash of our point is negative, then that point is a maximum. Now, letβ€²s find 𝑓 double dash of π‘₯.

In order to do this, we simply differentiate 𝑓 dash of π‘₯. And we found earlier that 𝑓 dash of π‘₯ is equal to three π‘₯ squared minus three. So 𝑓 double dash of π‘₯ is equal to d by dπ‘₯ of three π‘₯ squared minus three. Here, we’re differentiating another polynomial. So we take each term, we multiply by the power, and decrease the power by one. The first term is three π‘₯ squared. So we multiply by the power which is two. Multiplying the two by the three which is already there, we get six. We then decrease the power by one to give us π‘₯ to the power of one and multiply by this. And π‘₯ to the power of one is simply π‘₯. So we end up with six π‘₯. The next term negative three is a constant. And when we differentiate a constant, we simply get zero. So we can ignore this here. And we found that 𝑓 double dash of π‘₯ is equal to six π‘₯.

Now, we simply substitute in our critical points to find out whether they’re maximum or minimum. Our first critical point is at π‘₯ is equal to one. And 𝑓 double dash of one is equal to six. And six is positive. Therefore, at π‘₯ equals one, we must have a minimum. Our next critical point is at negative one. So we find 𝑓 double dash of negative one. And this is equal to negative six. And negative six is negative. So therefore, at π‘₯ equals negative one, we must have a maximum.

Now, we have found the π‘₯-values of our local minimum and maximum. We just need to find the value of the function at these π‘₯-values. To do this, we simply substitute π‘₯ is equal to one or minus one into 𝑓 of π‘₯. 𝑓 of one is equal to one cubed minus three times one minus two which is also equal to one minus three minus two or negative four. 𝑓 of negative one is equal to negative one cubed minus three times negative one minus two which is also equal to negative one plus three minus two or zero. Now, we have found that we have a local minimum at π‘₯ is equal to one and 𝑓 of π‘₯ is equal to negative four and a local maximum at π‘₯ is equal to negative one, 𝑓 of π‘₯ is equal to zero. This is the solution to the first part of this question.

Now, we need to find the inflection point of the curve of the function, if it exists. Letβ€²s think about what an inflection point is. An inflection point is where the curve changes between being convex upwards and convex downwards. Another way of thinking about this is a point of inflection is a point at which a change in the direction of the curvature occurs. In mathematical terms, it is a point where the second derivative of a function is zero or 𝑓 double dash of π‘₯ is equal to zero.

We found 𝑓 double dash of π‘₯ is equal to six π‘₯ earlier. So in order to find our point of inflection-in order to find our inflection point, we simply set 𝑓 double dash of π‘₯ which is equal to six π‘₯ to zero. So we have that six π‘₯ is equal to zero. And this gives us that π‘₯ is equal to zero. Now, we simply substitute π‘₯ equals zero into 𝑓 of π‘₯ and find that 𝑓 of zero is equal to zero cubed minus three times zero minus two, which is also equal to negative two.

Now, we have found that our inflection point is at π‘₯ is equal to zero and 𝑓 of π‘₯ is equal to negative two.

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