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.

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

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