Answer
For $a=3$ and $b=-3/2$, $f(x)$ is differentiable for all $x$-values.
Work Step by Step
$f(x)=ax+b$ for $x\gt-1$ and $f(x)=bx^2-3$ for $x\le-1$
1) For all $x\gt-1$: $$f'(x)=(ax+b)'=a\times1+0=a$$
Since $a$ is a definite value, $f(x)$ is differentiable on $(-1,\infty)$.
2) For all $x\lt-1$: $$f'(x)=(bx^2-3)'=2bx-0=2bx$$
For all $x\lt-1$, $(2bx)$ always gives out a definite value, so $f(x)$ is differentiable on $(-\infty,-1)$.
3) At $x=-1$:
First, for $f(x)$ to be differentiable at $x=-1$, $f(x)$ must be continuous at $x=-1$., which requires that $\lim_{x\to-1}f(x)=f(-1)$, or we might even say, $\lim_{x\to-1^+}f(x)=\lim_{x\to-1^-}f(x)=f(-1)$
We have $$f(1)=b\times1^2-3=b-3$$ $$\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}(ax+b)=a(-1)+b=b-a$$ $$\lim_{x\to-1^-}f(x)=\lim_{x\to-1^-}(bx^2-3)=b(-1)^2-3=b-3$$
Therefore, for $f(x)$ to be continuous at $x=-1$, we need $$b-a=b-3$$ $$-a=-3$$ $$a=3$$
Now we find the derivatives at $x=-1$:
- Left-hand derivative: $f'(-1)=2b\times(-1)=-2b$
- Right-hand derivative: $f'(-1)=a=3$
For $f(x)$ to be differentiable at $x=-1$, the left-hand derivative must be equal with the right-hand derivative. That means
$$-2b=3$$ $$b=-\frac{3}{2}$$
Therefore, for $a=3$ and $b=-3/2$, $f(x)$ is differentiable for all $x$-values.
