the derivative function

Printable View

October 15th 2008, 07:27 AM
vinson24

the 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
October 15th 2008, 09:17 AM
Chris L T521

Quote:

Originally Posted by vinson24

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?

A function is continuous at a point when $\lim_{x\to c}f(x)$ exists. This means that $\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)$

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

It is differentiable at that point if the derivatives of the first and second terms in the piecewise function are the same at that point.

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
October 15th 2008, 09:36 AM
vinson24

im still not seeing it
October 15th 2008, 10:38 AM
vinson24

i know f is continuous if k=1/5 and f is differentianle if k=3/80 but not seeing how to get there
October 15th 2008, 02:00 PM
Chris L T521

Quote:

Originally Posted by vinson24

i know f is continuous if k=1/5 and f is differentianle if k=3/80 but not seeing how to get there

Quote:

Originally Posted by vinson24

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?

Let's find $\lim_{x\to 5^-}f(x)$

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

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

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

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

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

Let's look at the derivative of the function for $x<5$ :

$f'(x)=2kx$

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

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

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

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

Does this make sense?

--Chris