1. derivative

suppose f(x)= sqrt of 1+3x, x>or equal to 5
k(x^2-5), x<5
for what value of k is continuous? For what value of k is f differentiable?

not even sure where to start i think i am on dealing with the second equation in this piecewise function

i know the answers are f is continuous if k=1/5 and f is differentiable if k=3/80

$f\left( x \right) = \left\{ \begin{gathered}
\sqrt {1 + 3x} ,{\text{ if }}x \geqslant 5 \hfill \\
k\left( {x^2 - 5} \right){\text{, if }}x < 5 \hfill \\
\end{gathered} \right. \hfill \\$

${\text{for }}f\left( x \right){\text{ to be continuous, it should be continuous at }}x = 5. \hfill \\$

${\text{For }}f\left( x \right){\text{ to be continuous at }}x = 5, \hfill \\$

$\mathop {\lim }\limits_{x \to 5 - } f\left( x \right) = \mathop {\lim }\limits_{x \to 5 + } f\left( x \right) = f\left( 5 \right) \hfill \\$

${\text{Now,}} \hfill \\$

$\mathop {\lim }\limits_{x \to 5 - } f\left( x \right) = \mathop {\lim }\limits_{x \to 5 - } k\left( {x^2 - 5} \right) = k\left( {5^2 - 5} \right) = 20k \hfill \\$

$\mathop {\lim }\limits_{x \to 5 + } f\left( x \right) = \mathop {\lim }\limits_{x \to 5 + } \sqrt {1 + 3x} = \sqrt {1 + 3\left( 5 \right)} = 4 \hfill \\$

$\Rightarrow 20k = 4 \hfill \\$

$\Rightarrow k = \frac{1}
{5} \hfill \\$

f(x) is continuous when k = 1/5

3. thanks how about finding the differentiable being 3/80

4. Originally Posted by vinson24
thanks how about finding the differentiable being 3/80
$f'\left( x \right) = \left\{ \begin{gathered}
\frac{3}{2\sqrt {1 + 3x}} \hfill \\
2kx \hfill \\
\end{gathered} \right. \hfill \\$

at x=5 , I think this is similar to before but using the above, you should look over any notes you have on continuity and differentiability.