Theory and Examples

In Exerc...

Question

Theory and Examples

In Exercises 51 and 52, give reasons for your answers.

Let f(x) = |x³ − 9x|.

a. Does f'(0) exist?

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says given the function F of X, which is equal to the absolute value of the quantity of X 4 minus 8 X in quantity, does F0 exist? We're given two possible choices as our answers. For choice A, we have yes, and for choice B, we have no. Now, in order to determine if F 0 exists, we're going to be using the limit definition of a derivative. So we call for that definition that we'll have F of 0, and that's going to be equal to the limit. As H approaches 0, of the quantity of F of 0 plus H. Minus F of 0 in quantity, divided by H. Now, in order for this derivative to exist, that means that the limit must also exist. And recall that for a limit to exist, one of the conditions is that the left-hand limit must approach the same value as the right hand limit. So we're going to calculate the left-hand limit and the right hand limit of our quotient here. In order to save time, we're going to go ahead and calculate F of 0 and F 0 plus H, which is just F of H, since it'll appear in both the left-hand limit and the right-hand limit. So, looking at F of 0, and evaluating the expression, that's going to be equal to the absolute value of the quantity of 0 to 4th, minus 8 multiplied by 0 in quantity, and evaluating this expression, this is going to be equal to 0. And evaluating F of H, that's going to be equal to the absolute value of the quantity of H the 4th minus 8 H in quantity. So, we'll first begin by looking at the left-hand limit of our quotient. So we'll have the limit as H approaches 0 from the left. Of the quantity, and here for F of H, if we look at the quantity inside the absolute value signs, we'll see that H the fourth minus 8H is going to be a positive quantity when H is slightly negative. And since H is approaching 0 from the left, it's going to take on slightly negative values as it approaches 0 from the left. So that means that we can just remove the absolute value signs from our function F of H, since the quantity inside the absolute value function is going to be positive. So, in the numerator for F of H, we're going to have H 4th, minus 8 H. Then we'll have minus F of 0, which is 0, and all of that is going to be divided by H. And we can simplify this expression by dividing each of the terms in the numerator by H. So we'll have the limit as H approaches 0 from the left of the quantity of H cubed minus 8 in quantity, and we can evaluate this limit by direct substitution, so this is going to be equal to. 0 cubed minus 8, which is equal to -8. We now need to look at the right hand limit, so we'll have the limit as H approaches 0 from the right of the quantity, and here for F of H, if we look at the quantity inside the absolute value sign, H the 4th minus 8 H is going to be slightly negative when H is slightly positive. And since we're approaching 0 from the right for H. That means that the quantity inside the absolute value of sign, is going to be actually a negative quantity. So to remove the absolute value signs, we'll need to multiply the quantity inside the absolute value signs by -1. So that means for F of H, we're going to have minus the quantity of H the 4th minus 8 H in quantity. Then we'll have minus F of 0, which is 0, and all of that is still going to be divided by H. Again, we can simplify this expression by dividing each of the terms of the numerator by H. So this is going to be equal to the limit. As H approaches 0 from the right of quantity of minus H cubed plus 8 in quantity. And evaluating this limit by direct substitution, this is going to be equal to minus 0 cubed. Plus 8, which is equal to 8, and here we see that our left hand limit does not equal the same value as our right hand limit. So that means that our limit does not exist. And since our limit does not exist, that means that our derivative F 0 also does not exist. And so that means that the answer to this problem is actually going to be no, and that corresponds to answer choice B. So I hope that this explanation has been useful, and I'll see you in the next video.

Theory and Examples

In Exercises 51 and 52, give reasons for your answers.

Let f(x) = |x³ − 9x|.

a. Does f'(0) exist?

Key Concepts

Derivative at a Point

Absolute Value Function

Piecewise Functions

Watch next