For $a>0$, $$f_a(x)=\begin{cases} 2ax^2+a(a-2)x+\frac{a^2}{2} &\text{ if $x<\frac{1}{2}$,}\\ 0 &\text{ if $x=\frac{1}{2}$,}\\ \dfrac{1-\sin(\pi x)}{\log(2x^2+\frac{1}{2})} &\text{ if $x>\frac{1}{2}$}. \end{cases}$$
- Determine for which $a$, $f_a$ is continuous at $x=1/2$.
- Determine for which $a$, $f_a$ is differentiable at $x=1/2$.
My attempt:
I calculated that $f_a$ is continous for $a=0 \ \lor \ a=\frac{1}{2}$ (Should be correct).
In order to solve $2.$, I calculated the limit of the difference quotient of both functions, and if I am not mistaken (I checked on Wolfram Alpha), $\nexists a$ for which $f_a$ is differentiable. Am I right?
You are correct. The function is continuous if and only if $$\lim_{x\to (1/2)^+}f_a(x)=\lim_{x\to (1/2)^+}\frac{1-\sin(\pi x)}{\log(2x^2+\frac{1}{2})}=0=f(1/2)=\lim_{x\to (1/2)^-}f_a(x)=a(a-1/2),$$ that is for $a=0$ or $a=1/2$. The right derivative at $x=1/2$ is $$\lim_{x\to (1/2)^+}\frac{f_a(x)-f_a(1/2)}{x-1/2}=\lim_{x\to (1/2)^+}\frac{1-\sin(\pi x)}{\log(2x^2+\frac{1}{2})(x-1/2)}=\frac{\pi^2}{4}.$$ As regards the left derivative, for $a=0$, $$\lim_{x\to (1/2)^-}\frac{f_a(x)-f_a(1/2)}{x-1/2}=\lim_{x\to (1/2)^-}\frac{0}{x-1/2}=0,$$ and, for $a=1/2$, $$\lim_{x\to (1/2)^-}\frac{f_a(x)-f_a(1/2)}{x-1/2}=\lim_{x\to (1/2)^-}\frac{x^2-3x/4+1/8}{x-1/2}=\frac{1}{4}.$$ Therefore, in both cases, the right derivative is not equal to left derivative and $f$ is not differentiable at $1/2$ for any value of $a$.
P.S. If we replace $\log(2x^2+\frac{1}{2})$ with $\pi^2\log(2x^2+\frac{1}{2})$ then $f_a$ is differentiable for $a=1/2$.