Video: APCALC03AB-P1A-Q12-848106168939

Let 𝑓 be the function given by 𝑓(π‘₯) = π‘₯³𝑒^(π‘Žπ‘₯Β²), where π‘Ž is a constant. For what value of π‘Ž does 𝑓 have a critical point at π‘₯ = 3/2?

04:50

Video Transcript

Let 𝑓 be the function given by 𝑓 of π‘₯ equals π‘₯ cubed 𝑒 to the power of π‘Žπ‘₯ squared, where π‘Ž is a constant. For what value of π‘Ž does 𝑓 have a critical point at π‘₯ equals three over two?

The critical points of a function are the points in its domain for which the first derivative of that function β€” 𝑓 prime of π‘₯ β€” is equal to zero or is undefined. We’re seeking the value of this constant π‘Ž for which our function has a critical point when π‘₯ is equal to three over two, which means we’re seeking the value of π‘Ž for which 𝑓 prime of three over two is either zero or undefined. We’re, therefore, going to need to find an expression for the first derivative 𝑓 prime of π‘₯ of our function.

Now looking at our function, we see that it is in fact a product of two differentiable functions: π‘₯ cubed and 𝑒 to the power of π‘Žπ‘₯ squared. And in order to find its derivative, we’ll need to apply the product rule. The product rule tells us that for two differentiable functions, 𝑒 and 𝑣, the derivative with respect to π‘₯ of their product, 𝑒𝑣, is equal to 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. We multiply each function by the derivative of the other and add them together. We, therefore, let 𝑒 equal our first function β€” that’s π‘₯ cubed β€” and 𝑣 equal the second function β€” that’s 𝑒 to the power of π‘Žπ‘₯ squared.

We then need to find each of their individual derivatives. To find the derivative of 𝑒 with respect to π‘₯, we will call the power rule of differentiation. And we see that d𝑒 by dπ‘₯ is equal to three π‘₯ squared. In order to find d𝑣 by dπ‘₯, we’re going to need to recall standard results for differentiating exponential functions. We recall that the derivative with respect to π‘₯ of 𝑒 to the power of some function, 𝑔 of π‘₯, is equal to 𝑔 prime of π‘₯ multiplied by 𝑒 to the power of 𝑔 of π‘₯, which is an application of the chain rule.

Here, our function 𝑔 of π‘₯ is π‘Žπ‘₯ squared and its derivative with respect to π‘₯ is two π‘Žπ‘₯, recalling that π‘Ž is a constant. So we have that d𝑣 by dπ‘₯ is equal to two π‘Žπ‘₯ multiplied by 𝑒 to the power of π‘Žπ‘₯ squared. We can now substitute into our formula for the product rule. 𝑓 prime of π‘₯ is equal to 𝑒 times d𝑣 by dπ‘₯. That’s π‘₯ cubed multiplied by two π‘Žπ‘₯𝑒 to the π‘Žπ‘₯ squared plus 𝑣 times d𝑒 by dπ‘₯. That’s 𝑒 to the power of π‘Žπ‘₯ squared multiplied by three π‘₯ squared. We can factor by π‘₯ squared 𝑒 to the power of π‘Žπ‘₯ squared. And we’re left with 𝑓 prime of π‘₯ is equal to π‘₯ squared 𝑒 to the π‘Žπ‘₯ squared multiplied by two π‘Žπ‘₯ squared plus three.

Now recall that we said that in order for this function to have a critical point at π‘₯ equals three over two, it must be the case that our first derivative is zero or is undefined at this point. Our function 𝑓 prime of π‘₯ doesn’t have any discontinuities. So it must be the case that the first derivative is zero when π‘₯ is equal to three over two. We can, therefore, substitute zero for 𝑓 prime of three over two and three over two for π‘₯ to give an equation that we can solve in order to find the value of π‘Ž.

Now notice that each of our first two factors are not equal to zero. Three over two squared is not equal to zero and 𝑒 to the power of π‘Ž three over two squared is not equal to zero, which means that we can divide both sides of our equation by these factors to give a simplified equation. Zero is equal to two π‘Ž times three over two squared plus three. Three over two squared is equal to nine over four. And then, we can cancel a common factor of two to leave us with π‘Ž multiplied by nine over two.

Our equation is, therefore, simplified to nine over two plus three is equal to zero. We can then subtract three from each side to give nine π‘Ž over two equals negative three and then multiply by two-ninths to eliminate the fraction on the left-hand side, giving π‘Ž equals negative three multiplied by two over nine. We can cancel a common factor of three from the numerator and denominator to give π‘Ž equals negative one multiplied by two over three. That’s just simplifies to negative two-thirds.

So by recalling that the critical points of a function occur when its first derivative is equal to zero or is undefined and by applying the product rule, we found that for this function, 𝑓 of π‘₯, it will have a critical point at π‘₯ equals three over two when π‘Ž is equal to negative two-thirds.

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