# find a and b such that piecewise this piecewise function is twice differentiable

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.

• Thank you! Why is it okay to just plug in x for t? Nov 12 '19 at 12:15
• wait no, i did it wrong HAHA... Nov 12 '19 at 13:25
• the answer doesn't change, but you are correct in pointing out my mistake. i have corrected it Nov 12 '19 at 13:26