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
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$\begingroup$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.
$\endgroup$ $\begingroup$$4+๐=8๐$ is correct. what should you get from this Eric?
Also $3๐๐ฅ^2=2๐ฅ$. But why Eric?
$\endgroup$ 2 $\begingroup$You made an error saying that
$$(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$$
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