Question

Please help, It's calculus work. The screenshot below should answer my question.
80 Points

Answer 1

a) In order for $f$ to be continuous at $x=1$, we need to have

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$

By definition of $f$, we know $f(1)=10-2-1^2=7$. Meanwhile, the limits are

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}10-2x-x^2=7$

$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}3+4e^{x-1}=7$

so $f$ is indeed continuous at $x=1$.

b) We use the first derivative test (FDT) here, but when we compute the derivative of a piecewise function, we have to be careful at the points where the pieces "split off", because it's possible that the derivative does not exist at these points, yet an extreme value can still occur there. (Consider, for example, $|x|$ at $x=0$.)

In this case,

$f'(x)=\begin{cases}-2-2x&\text{for }x<1\\?&\text{for }x=1\\4e^{x-1}&\text{for }x>1\end{cases}$

We find the critical points for each piece over their respective domains:

On the first piece:

$-2-2x=0\implies x=-1$

which does fall in [-2, 2]. The FDT shows $f'(x)<0$ for $x$ less than and near -1, and $f'(x)>0$ for $x$ greater than and near -1, so $f(-1)=11$ is a local maximum.

On the second piece:

$4e^{x-1}\neq0\text{ for all }x$

so it does not contribute any critical points.

Where the pieces meet:

By checking the conditions for continuity mentioned in part (a), we can determine that $f'(1)$ does not exist, but that doesn't rule out $x=1$ as a potential critical point.

We have

$\displaystyle\lim_{x\to1}-2-2x=-4$

so $f'(x)<0$ for $x$ less than and near 1, and

$\displaystyle\lim_{x\to1}4e^{x-1}=4$

so $f'(x)>0$ for $x$ greater than and near 1. So the FDT tells us that $f(1)=7$ is a local minimum.

Finally, at the endpoints of the domain we're concerned with, [-2, 2]:

We have $f(-2)=10$ and $f(2)=3+4e$.

So, on [-2, 2], $f$ attains an absolute minimum of $f(1)=7$ and an absolute maximum of $f(2)=3+4e\approx13.9$.