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

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- Oct 15th 2008, 07:27 AMvinson24the derivative function
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 - Oct 15th 2008, 09:17 AMChris L T521
A function is continuous at a point when $\displaystyle \lim_{x\to c}f(x)$ exists. This means that $\displaystyle \lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)$

Quote:

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

Note that you may not get the same k value for the continuity and differentiability cases...

Can you try these now and see what you get?

--Chris - Oct 15th 2008, 09:36 AMvinson24
im still not seeing it

- Oct 15th 2008, 10:38 AMvinson24
i know f is continuous if k=1/5 and f is differentianle if k=3/80 but not seeing how to get there

- Oct 15th 2008, 02:00 PMChris L T521
Let's find $\displaystyle \lim_{x\to 5^-}f(x)$

$\displaystyle \lim_{x\to 5^-}f(x)=\lim_{x\to 5} k(x^2-5)=20k$

Now find $\displaystyle \lim_{x\to 5^+}f(x)$

$\displaystyle \lim_{x\to 5^+}f(x)=\lim_{x\to 5} \sqrt{1+3x}=\sqrt{16}=4$

The function is continuous when $\displaystyle \lim_{x\to 5^-}f(x)=\lim_{x\to 5^+}f(x)\implies 20k=4\implies \color{red}\boxed{k=\tfrac{1}{5}}$

Does this bit make sense?

Quote:

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

$\displaystyle f'(x)=2kx$

Now, let's look at the derivative of the function for $\displaystyle x\geq5$:

$\displaystyle f'(x)=\frac{3}{2\sqrt{1+3x}}$

In order for this function to be differentiable at $\displaystyle x=5$, both of these derivatives must have the same value at $\displaystyle x=5$:

This implies that $\displaystyle 10k=\frac{3}{8}\implies \color{red}\boxed{k=\frac{3}{80}}$

Does this make sense?

--Chris