Question

consider the following function.
f(x)=\begin{cases}5x^{2}+3x + 10&xleq1\ax + 5&x>1end{cases}
determine a value of a such that the function is differentiable everywhere. give an exact answer.
otherwise, enter dne if no such value exists.
(a=)

Explanation:

Step1: Check continuity at x = 1

For a function to be differentiable everywhere, it must be continuous. So, $\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)$.
$\lim_{x
ightarrow1^{-}}(5x^{2}+3x + 10)=5(1)^{2}+3(1)+10=5 + 3+10 = 18$.
$\lim_{x
ightarrow1^{+}}(ax + 5)=a(1)+5=a + 5$.
Set $a + 5=18$, so $a=13$ for continuity.

Step2: Check differentiability at x = 1

Find the derivative of each part.
For $y = 5x^{2}+3x + 10$ ($x\leq1$), $y^\prime=10x+3$. At $x = 1$, $y^\prime(1)=10(1)+3 = 13$.
For $y=ax + 5$ ($x>1$), $y^\prime=a$.
Since the function is differentiable at $x = 1$, the left - hand derivative and right - hand derivative must be equal. When $a = 13$, the right - hand derivative is also 13.

Answer:

$13$