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In Exercises 115 and 116, find [tex]a[/tex] and [tex]b[/tex] such that [tex]f[/tex] is differentiable everywhere.

115.

\[
f(x) = \begin{cases}
a x^3, & x \leq 2 \\
x^2 + b, & x > 2
\end{cases}
\]

Answer :

To make the function [tex]\( f(x) \)[/tex] differentiable everywhere, we need to ensure that both the function and its derivative are continuous at [tex]\( x = 2 \)[/tex].

The function is given as:
[tex]\[
f(x) = \begin{cases}
a x^3 & \text{if } x \leq 2 \\
x^2 + b & \text{if } x > 2
\end{cases}
\][/tex]

### Step 1: Ensure the Function is Continuous at [tex]\( x = 2 \)[/tex]
First, we set the two pieces of the function equal to each other at [tex]\( x = 2 \)[/tex].

For [tex]\( x \leq 2 \)[/tex]:

[tex]\[
f(2) = a \cdot 2^3 = 8a
\][/tex]

For [tex]\( x > 2 \)[/tex]:

[tex]\[
f(2) = 2^2 + b = 4 + b
\][/tex]

Set the expressions equal to ensure continuity at [tex]\( x = 2 \)[/tex]:

[tex]\[
8a = 4 + b
\][/tex]

### Step 2: Ensure the Derivative is Continuous at [tex]\( x = 2 \)[/tex]
Now, we need to find the derivatives of the two pieces and set them equal at [tex]\( x = 2 \)[/tex].

For [tex]\( x \leq 2 \)[/tex]:

[tex]\[
f'(x) = \frac{d}{dx} (a x^3) = 3a x^2
\][/tex]

Evaluating at [tex]\( x = 2 \)[/tex]:

[tex]\[
f'(2) = 3a \cdot 2^2 = 12a
\][/tex]

For [tex]\( x > 2 \)[/tex]:

[tex]\[
f'(x) = \frac{d}{dx} (x^2 + b) = 2x
\][/tex]

Evaluating at [tex]\( x = 2 \)[/tex]:

[tex]\[
f'(2) = 2 \cdot 2 = 4
\][/tex]

Set the derivatives equal to ensure differentiability at [tex]\( x = 2 \)[/tex]:

[tex]\[
12a = 4
\][/tex]

Solve for [tex]\( a \)[/tex]:

[tex]\[
a = \frac{4}{12} = \frac{1}{3}
\][/tex]

### Step 3: Solve for [tex]\( b \)[/tex]
Using the value of [tex]\( a \)[/tex] in the continuity equation:

[tex]\[
8a = 4 + b
\][/tex]

Substitute [tex]\( a = \frac{1}{3} \)[/tex]:

[tex]\[
8 \cdot \frac{1}{3} = 4 + b \implies \frac{8}{3} = 4 + b
\][/tex]

Now solve for [tex]\( b \)[/tex]:

[tex]\[
b = \frac{8}{3} - 4 = \frac{8}{3} - \frac{12}{3} = -\frac{4}{3}
\][/tex]

### Solution
The values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] that make the function [tex]\( f(x) \)[/tex] differentiable everywhere are:

[tex]\[
a = \frac{1}{3}, \quad b = -\frac{4}{3}
\][/tex]

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