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
You only have problems to differentiate at $\displaystyle x=2$ so you must find $\displaystyle a,b$ such as $\displaystyle f$ is continuous and differentiable there, i.e. to find values such as $\displaystyle \lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{+}}f(x)=f(2)$ and $\displaystyle \lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$ exists.
$\displaystyle \lim_{x\to 2^{+}}f(x)=\lim_{x\to 2^{+}}x^2+b=4+b$ and $\displaystyle f(2)=8a$ so our first condition is $\displaystyle 4+b=8a$.
On the other hand, $\displaystyle \lim_{h\to 0^{+}}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^{+}}\frac{(2+h)^2+b-4-b}{h}=\lim_{h\to 0^{+}}\frac{(2+h)^2-4}{h}=4.$ and $\displaystyle \lim_{h\to 0^{-}}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^{-}}\frac{a(2+h)^3-8a}{h}=4a$ so our second condition is $\displaystyle 4a=4$.
Finally $\displaystyle a=1$ and $\displaystyle b=4$.