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
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
A function is continuous at a point when $\displaystyle \lim_{x\to c}f(x)$ exists. This means that $\displaystyle \lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)$
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.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
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?
Let's look at the derivative of the function for $\displaystyle x<5$: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