Find all triples $(a,b,c)$ such that $$f(x)=ax^2+b ,x\le1$$$$f(x)=cx^4-2x^2,x>1$$ is differentiable at 1.
I assumed that I was supposed to use the limit definition of the derivative somehow, but this is giving me trouble since the function is piecewise and there are so many different possibilities for $(a, b, c)$. I'm honestly not really sure how to approach the problem; does anyone have any advice?
For a function to be differentiable it must be continuous.
$\implies a+b = c-2$
and
$f'(1^-) = f'(1^+) \implies 2a = 4c-4\implies a = 2c-2$
substituing $a=2c-2$ into $a+b = c-2$
$2c-2 + b = c-2 \implies b = -c$
$(2c-2,-c,c)$ is the required triple.