Question Video: Checking the Value of an Absolute Minimum | Nagwa Question Video: Checking the Value of an Absolute Minimum | Nagwa

# Question Video: Checking the Value of an Absolute Minimum Mathematics

True or False: The function π(π₯) = 2(π₯Β² β 1)Β³ has an absolute minimum over the interval [β1, 1] that is equal to β2.

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### Video Transcript

True or False: The function π of π₯ is equal to two times π₯ squared minus one all cubed has an absolute minimum over the closed interval from negative one to one that is equal to negative two.

In this question, we need to determine whether a statement is true or false involving the absolute minimum value of a function on a closed interval. And to do this, we can start by checking if our function even has an absolute minimum value on this interval. And to answer this question, weβll recall part of the extreme value theorem. This tells us that continuous real-valued functions on closed bounded intervals will achieve a maximum and a minimum value. We only need the minimum value part. We can apply this directly to our function π of π₯. First, we can see that π of π₯ is two multiplied by a quadratic all cubed. In other words, π of π₯ is a polynomial.

And we recall polynomials are continuous for all real values of π₯. Next, we can see weβre working over the closed interval from negative one. This is a closed bounded interval. Therefore, π of π₯ is a continuous function on a closed bounded interval. We can apply the extreme value theorem, which tells us it hits an absolute minimum value on this interval. But weβre not done yet. To conclude that this statement is true, we need to check that this value is equal to negative two.

To do this, we need to recall how we find the absolute maximum and minimum values of a function on a closed interval. We do this in three steps. First, we find all the critical points of our function π of π₯ on the closed interval. And we recall these are the points where π prime of π₯ is equal to zero or where the derivative does not exist.

And we find these values because, remember, the local extrema of a function are always at the critical points. And itβs worth pointing out in this case because our function π of π₯ is a polynomial, we know itβs differentiable for all real values of π₯. So the only critical points of our function will be where the derivative is equal to zero. Second, we evaluate π at the critical points. This will give us the possible values of local extrema. And, finally, we evaluate π at the endpoints of our interval because these become absolute extrema over the closed interval.

So letβs start with the first step. We need to find all the critical points of our function. Remember, in this case, these are only going to be when its derivative is zero. Thereβs a few different ways of doing this. Because we have a composition of two functions, we could use the chain rule or the general power rule. However, in this case, the exponent value of three is quite low. So we can just expand this by using the binomial theorem or the FOIL method. And any of these methods will work. Weβll use the general power rule.

And we recall the general power rule is an application of the chain rule. It tells us the derivative of π of π₯ all raised to the power of π is equal to π times π prime of π₯ multiplied by π of π₯ all raised to the power of π minus one. And this is provided π is differentiable at π₯ and both sides of the equation are well defined. And we want to apply this to differentiate π of π₯. Remember, π of π₯ is a polynomial. So we already know itβs differentiable for all values of π₯. So we set our exponent value of π equal to three and our inner function π of π₯ equal to π₯ squared minus one.

We can just take the constant factor of two outside of our derivative. Now, all we need to do to differentiate π of π₯ is find an expression for π prime of π₯. And we can find an expression for π prime of π₯ by using the power rule for differentiation term by term. We multiply by the exponent of π₯ and reduce this exponent by one. π prime of π₯ is π₯ squared. Now, we can find π prime of π₯ by substituting our expressions for π of π₯, π prime of π₯, and π into the general power rule. Remember, we need to multiply this by a constant factor of two. We get π prime of π₯ is two times three π₯ squared multiplied by π₯ squared minus one raised to the power of three minus one.

And we can simplify this. Two times three π₯ squared is six π₯ squared and three minus one is two. So π prime of π₯ is six π₯ squared multiplied by π₯ squared minus one all squared. And now we can use this to find the critical points of our function. This will be where the derivative is zero. Therefore, we need to solve the equation six π₯ squared multiplied by π₯ squared minus one all squared is equal to zero. And to do this, we recall if the product of two terms is equal to zero, then one of the two factors must be equal to zero. Therefore, for the derivative of π to be equal to zero, either six π₯ squared is equal to zero or π₯ squared minus one all squared is equal to zero.

We can then solve both of these equations separately. First, if six π₯ squared is equal to zero, then π₯ squared is equal to zero. So π₯ must be equal to zero. Next, since π₯ squared minus one all squared is zero, we must have the base π₯ squared minus one is zero, which we can then solve by adding one to both sides of the equation and taking the square root, where we get a positive and a negative root. π₯ is either positive or negative one. Therefore, weβve shown that our function π has critical points when π₯ is equal to zero, one, and negative one. This is all the possible critical points.

We can now move on to the second step and evaluate our function at these three values. Letβs start when π₯ is equal to zero. We substitute zero into π of π₯ to get two times zero squared minus one all cubed, which if we evaluate is equal to negative two. We can do the same to evaluate π at one. W get two times one squared minus one all cubed, which has a factor of zero. So π evaluated at one is equal to zero. And finally, we can evaluate π at negative one. We get two times negative one all squared minus one all cubed. And this also has a factor of zero, so π evaluated at negative one is zero. And now that weβve evaluated π at the critical points, we can move on to evaluating π at the endpoints of our interval. Thatβs negative one and one.

However, in this case, we can already see that the endpoints of this interval are critical points of the function. So weβve already evaluated π at the endpoints of our interval. Now, we can conclude the largest of these values will be the absolute maximum of our function on the closed interval from negative one to one. And the lowest of these three values will be the absolute minimum of our function on this interval. We can see this is negative two. Therefore, we were able to show itβs true that the function π of π₯ is equal to two times π₯ squared minus one all cubed has an absolute minimum value of negative two over the closed interval from negative one to one.