I am stuck with this homework problem..
Given the following piecewise function: \begin{cases} y = \int_0^{2x} \sqrt{1+t^4} dt, & x\gt 0 \\ y = ax+bx^2 & x \le 0 \ \end{cases}
Find $a$ and $b$ such that it is twice differentiable. All help is greatly appreciated...
My idea of an approach here is to make a set of equations by checking for continuity at zero, which gives an equation, then calculating the derivative at zero, which gives another equation, then solve this set of equations. However, I don't know what to do with that integral..
I'd love to show some work, and I have been idling on how to start before posting this..If you feel like not doing the work (as expected), please give me a direction:))
First, we check that the function is continuous.
Yes, $\displaystyle \lim_{x \to 0^+}\int_0^{2x}\sqrt{1+t^4}\,dt=\lim_{x\to0^-}ax+bx^2=0$.
Now, we take the derivative of each side, and reevaluate the limit.
We need that $\displaystyle \lim_{x \to 0^+}2\sqrt{1+(2x)^4}=\lim_{x \to 0^-}a+2bx=2$.
That's our first equation: $a=2$.
Then, we take the second derivative, and reevaluate.
We have that $\displaystyle \lim_{x \to 0^+}\frac{48x^3}{\sqrt{1+16x^4}}=\lim_{x \to 0^-}2b=0$.
So for $a=2$ and $b=0$, the function is twice differentiable.