Find $a$ and $b$ so that $f$ is differentiable everywhere

Question

The piecewise function is $$f(x)=\begin{cases} ax^3 & x\leq 2 \\ x^2+b, & x>2 \end{cases} $$

First I plug in the 2 into both functions and set them equal to each other $$ 2^2+b=a(2)^3$$ $$4+b=8a$$ $$\frac{1}{2}=a-b $$ Then I take the derivative of both functions and repeat $$3ax^2=2x+b$$ $$12a=4+b$$ $$a+b=\frac{1}{3} $$

I then subtract both equations from each others $$\frac{1}{2}=a-b $$ $$-\frac{1}{3}=a+b$$ to get $b=\frac{1}{6}$ and then $a=\frac{2}{3}$ but when I checked my answer I was wrong, I don't know where I went wrong

Answer 1

You are on the right track but you have made some algebra and derivative mistake in finding your $a$ and $b$.

Note that $$12a=4 \implies a=1/3$$

Also $$ 4+b=8a \implies a=1/2 +b/8$$

You can take it from here.

Answer 2

$4+=8$ is correct. what should you get from this Eric?

Also $3^2=2$. But why Eric?

Answer 3

You made an error saying that

$$(x^2+b)' = 2x+b$$

The correct way to get the value of $b$: Function $f(x)$ must be continuous at $x=2$. You will get:

$$8a = 4+b \quad (1)$$

And the derivative must also be continuous:

$$12a = 4$$