The right-sided and left-sided derivatives of a function at a ...

Question

The right-sided and left-sided derivatives of a function at a point $a$ are given by $f_{+}^{'} (a) = \lim_{h \to 0^{+}} \frac{f (a + h) - f (a)}{h}$ and $f_{-}^{'} (a) = \lim_{h \to 0^{-}} \frac{f (a + h) - f (a)}{h}$ , respectively, provided these limits exist. The derivative $f^{\prime}\left(a\right)$ exists if and only if $f_{+}^{\prime}\left(a\right)=f_{-}^{\prime}\left(a\right)$ .
Compute $f_{+}^{\prime}\left(a\right)$ and $f_{-}^{\prime}\left(a\right)$ at the given point $a$ .
$f (x) = {\begin{matrix} 4 - x^{2} if x \leq 1 \\ 2 x + 1 if x > 1 \end{matrix}$ ; $a equals 1$

Compute $f_{+}^{\prime}\left(a ight)$ and $f_{-}^{\prime}\left(a ight)$ at the given point $a$ .

$f (x) = {\begin{matrix} 4 - x^{2} if x \leq 1 \\ 2 x + 1 if x > 1 \end{matrix}$ ; $a equals 1$

Accepted Answer

Hi everyone, let's take a look at this practice problem dealing with derivatives. This problem says the following formulas for F minus of A and F+ of A represent the left and right-sided derivatives of a function at a point A, respectively. And we're given F minus of A. is equal to the limit as H approaches 0 from the left of the quantity of F of A plus H minus F of A in quantity divided by H and F+ of A is equal to the limit as H approaches 0 from the right of the quantity of F of A plus H minus F of A in quantity divided by H. We're asked to consider the function F of X, which is equal to 6 minus X2d if X is less than or equal to 2, and 3x minus 4 if X is greater than 2, and we're asked to find F minus of A and F+ of A at A equal to 2. We're given 4 possible choices as our answers. For choice A, we have F minus 2 is equal to -4. And F + 2 is equal to 3. For choice B, we have F minus 2 is equal to -4, and F + 2 is equal to minus 3. For choice C we have F minus 2 is equal to 6, and F + 2 is equal to 3. And for choice D we have F minus 2 is equal to 6, and F + 2 is equal to minus 3. Now, we're asked to find F minus A and F+ of A at A equal to 2. So we're just going to use the formulas that we were given in the problem to calculate these quantities. So we're gonna start off by calculating F minus of A, and here A is going to be equal to 2, so we're going to be calculating F minus 2. And this is going to be equal to the limit. As H approaches 0 from the left of the quantity of F of 2 + H, Minus F of 2 in quantity divided by H. Now, in this case, we're taking the limit as H approaches 0 from the left. That means H is going to be slightly negative. And since H is slightly negative, the quantity of 2 plus H is going to be slightly less than 2. So that means we're going to want to use the first part of our piece-wise function. In order to calculate F of X, since X is going to be less than or equal to 2. So that means F of X is going to be equal to 6 minus X2. So substituting that into our expression here, this is going to be equal to the limit. As H approaches 0 from the left of the quantity, and for F of 2 + H, we're going to replace X with the quantity of 2 + H. So we'll have 6 minus the quantity of 2 + H in quantity squared. Minus the quantity, and here for F of 2, we replace X with 2, so that's gonna be 6 minus 2 squared. In quantity and all that is divided by H. So simplifying our numerator, this is going to be equal to the limit, as H approaches 0 from the left of the quantity of 6 minus and here we'll multiply out the quantity of 2 + H in quantity squared. That's gonna be the quantity of 4 + 4 H + H2. In quantity, and then we'll have minus, and we'll have 6 minus 4, which is 2. And all of that is still divided by H. So distributing our minus sign now into our parenthesis, this is going to be equal to the limit. As H approaches 0 from the left of the quantity of 6 minus 4, minus 4 H minus H squared, minus 2, and all that is divided by H. So now simplifying our numerator, we'll see that we have 6 minus 4 minus 2. I want to add that together, that gives me 0. The remaining terms all have H in them, so I can divide out one factor of H in the numerator for each of those terms with the H in the denominator. Doing so, this is going to be equal to the limit. As H approaches 0 from the left. Of the quantity of -4 minus H. So now we can evaluate this limit by direct substitution. So setting H equal to 0, this is going to be equal to minus 4. And so that is the first half of the answer for this question. So, F minus 2 is equal to -4. Now we're gonna do basically a similar calculation for F+ of A. And so for F+ of A, I'll be looking at F + 2. And this is going to be equal to the limit. As H approaches 0 from the right of the quantity of F of 2 + H. Minus F of 2 in quantity divided by H. Now, here, when H is approaching 0 from the right, H is going to be slightly positive. So, the quantity of 2 plus H is going to be slightly greater than 2. So that means I'm going to want to use the second half of our piece-wise function for F of X. In order to evaluate this limit. So, I'm going to want to use 4 F of X, 3X minus 4. So making that substitution, This is going to be equal to. The limit As H approaches 0 from the right. Of the quantity, and here for F of 2 + H, we're going to replace X in our equation with the quantity of 2 + H. So we'll have 3, multiplying the quantity of 2 + H in quantity minus 4. Then we're gonna have minus F of 2. However, In this case, we're gonna be looking for F of X when X is equal to 2, so we're gonna have to still use our first piece of our piece-wise function, since that was the case when X was less than or equal to 2. So this will be minus the quantity. Of 6 minus 2 squad. That's the same as our first calculation, or F minus 2. So, all of this is still divided by H. So, simplifying this expression, this is going to be equal to the limit. As H approaches 0 from the right. Of the quantity, and we'll distribute our 3, so we'll have 6 + 3 H. -4. -2 6 minus 2 squad gets me 2. And this whole quantity is still divided by H. Again, simplifying will have 6 minus 4 minus 2, which add up to 0. And we'll have 3 H divided by H, so I can cancel out the H's, so this is going to be equal to the limit, as H approaches 0 from the right. Of 3. So here we're taking the limit of a constant, and so anytime I take the limit of a constant, I get back that just that constant. So this is equal to just 3. And so that is the second half of the problem answered. So F + 2 is equal to 3. So now, if we take our answers that we've calculated and compare them to our answer choices, the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Right-sided and Left-sided Derivatives

Existence of the Derivative

Piecewise Functions

Watch next