Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part B β€’ Question 83

If 𝑔 is defined by 𝑔(π‘₯) = 3π‘₯ + 2𝑒^(βˆ’3π‘₯Β²) on the interval [0, 5], at which of the following values of π‘₯ does 𝑔 have a point of inflection? [A] βˆ’0.408 [B] 3.464 [C] 0.408 [D] 0.289

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Video Transcript

If 𝑔 is defined by 𝑔 of π‘₯ equals three π‘₯ plus two 𝑒 to the negative three π‘₯ squared on the closed interval zero to five, at which of the following values of π‘₯ does 𝑔 have a point of inflection? Is it a) negative 0.408, b) 3.464, c) 0.408, or d) 0.289?

Remember, a point of inflection is a point on the graph at which the concavity changes. In other words, the graph goes from being concave up to concave down, or vice versa. These points of inflection occur when the second derivative of our function is equal to zero. So in this case, we can say we need to find the points where 𝑔 double prime of π‘₯ is equal to zero.

Now we do need to be a little bit careful here. This doesn’t guarantee we found a point of inflection. So once we found the values for π‘₯ for which 𝑔 double prime of π‘₯ is equal to zero, we evaluate 𝑔 double prime of π‘₯ to either side of this point. We need to check that the second derivative is going from being a positive number to a negative number, or vice versa. So let’s find the second derivative of our function.

We’ll begin of course by finding the first derivative of three π‘₯ plus two 𝑒 to the negative three π‘₯ squared. The derivative of three π‘₯ with respect to π‘₯ is just three. But what about the derivative of two 𝑒 to the negative three π‘₯ squared? This is a function of a function. It’s a composite function. So we’re going to use the chain rule to find the derivative. This says that if 𝑦 is a function in 𝑒 and 𝑒 is a function in π‘₯, then d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯.

We’ll let therefore 𝑒 be equal to negative three π‘₯ squared. So 𝑦 is equal to two 𝑒 to the power of 𝑒. The derivative of negative three π‘₯ squared is two times negative three π‘₯. That’s negative six π‘₯. And d𝑦 by d𝑒 is just two 𝑒 to the power of 𝑒. So d𝑦 by dπ‘₯ is the product of these two. It’s negative six π‘₯ times two 𝑒 to the 𝑒, which is negative 12π‘₯𝑒 to the 𝑒 or negative 12π‘₯𝑒 to the negative three π‘₯ squared. And we can see that the first derivative 𝑔 prime of π‘₯ is three minus 12π‘₯𝑒 to the negative three π‘₯ squared.

We’re now going to find the second derivative. The derivative of three is zero. So we’re actually going to differentiate negative 12π‘₯𝑒 to the negative three π‘₯ squared with respect to π‘₯. This time, we have the product of two functions. So we’re going to use the product rule. This says that if 𝑒 and 𝑣 are differentiable functions, then the derivative of their product is 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯.

Now since we’re using 𝑒 in the chain rule, I’m going to change our functions to 𝑝 and π‘ž. We’ll let 𝑝 be equal to negative 12π‘₯ and π‘ž be equal to 𝑒 to the negative three π‘₯ squared. This time, d𝑝 by dπ‘₯ is equal to negative 12. And we could use the chain rule again to find the derivative of 𝑒 to the negative three π‘₯ squared. But we saw that the derivative of two 𝑒 to the negative three π‘₯ squared is negative 12π‘₯𝑒 to the negative three π‘₯ squared. So the derivative of 𝑒 to the negative three π‘₯ squared is half of this. It’s negative six π‘₯ times 𝑒 to the negative three π‘₯ squared. We substitute everything into our formula for the product rule. And when we simplify, we see that the second derivative is 72π‘₯ squared 𝑒 to the negative three π‘₯ squared minus 12𝑒 to the negative three π‘₯ squared.

Remember, we’re looking to find a point of inflection. So we’re gonna set this equal to zero and solve for π‘₯. We can factor 12𝑒 to the negative three π‘₯ squared. And then we see that, for this to be equal to zero, either 12𝑒 to the negative three π‘₯ squared must be equal to zero or six π‘₯ squared minus one must be equal to zero.

Now there are no values for π‘₯ for which 12𝑒 to the negative three π‘₯ squared will be equal to zero. So we know that six π‘₯ squared minus one must be equal to zero. We solve by adding one to both sides and then dividing through by six. We then take the square root of both sides of our equation, remembering we’ll need to find both the positive and negative square root of one-sixth.

Correct to three decimal places, we see that 𝑔 double prime of π‘₯ is equal to zero when π‘₯ is equal to 0.408 and negative 0.408. Remember, we said that this didn’t guarantee a point of inflection. So we need to find the value of the second derivative at points either side of 0.408 and negative 0.408.

Let’s begin by finding 𝑔 double prime of 0.4 and 0.41. 𝑔 double prime of 0.4 gives us a negative value. That’s concave down. And 𝑔 double prime of 0.41 gives us a positive value. That’s concave up. So there is indeed a point of inflection at this point, 0.408. Let’s check negative 0.408. This time, the second derivative at negative 0.41 and negative 0.4 is negative. So it’s concave down either side of the point π‘₯ equals negative 0.408. So this is not a point of inflection. The point of inflection occurs when π‘₯ is equal to 0.408.

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