# Thread: stuck on this one problem

1. ## stuck on this one problem

Can anyone help me out with this one problem please, didn't really learn this type of problem yet.

find a and b such that f is differentiable everywhere

f(x)= ax^3, x=less than or equal to 2
x^2+b, x=greater than 2

thanks

2. You only have problems to differentiate at $x=2$ so you must find $a,b$ such as $f$ is continuous and differentiable there, i.e. to find values such as $\lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{+}}f(x)=f(2)$ and $\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$ exists.

$\lim_{x\to 2^{+}}f(x)=\lim_{x\to 2^{+}}x^2+b=4+b$ and $f(2)=8a$ so our first condition is $4+b=8a$.

On the other hand, $\lim_{h\to 0^{+}}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^{+}}\frac{(2+h)^2+b-4-b}{h}=\lim_{h\to 0^{+}}\frac{(2+h)^2-4}{h}=4.$ and $\lim_{h\to 0^{-}}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^{-}}\frac{a(2+h)^3-8a}{h}=4a$ so our second condition is $4a=4$.

Finally $a=1$ and $b=4$.

3. thanks for showing me how to do it. just one question, could there be another answer because in my text book it is different than what you gave me.

4. Originally Posted by soldatik21
Can anyone help me out with this one problem please, didn't really learn this type of problem yet.

find a and b such that f is differentiable everywhere

f(x)= ax^3, x=less than or equal to 2
x^2+b, x=greater than 2

thanks
Continuity at x = 2: a(2)^3 = (2)^2 + b => 8a = 4 + b .... (A)

Differentiability at x =2: 3a(2)^2 = 2(2) => 12a = 4 => a = 1/3.

Substitute this value of a into equation (A) to get b.