Find the minimum and maximum of an exponential function on a closed interval

Key concept: Because the base is $3>1$, the function increases wherever the exponent increases. That means you do not need to differentiate the whole exponential expression if you can first study the quadratic in the exponent.

Step by step

We have

$f(x)=3^{x^2+2x+2},\qquad -2\le x\le 2.$

Since $3>1$, minimizing or maximizing $f(x)$ is the same as minimizing or maximizing the exponent

$g(x)=x^2+2x+2=(x+1)^2+1.$

Step 1: Find the minimum

Because $(x+1)^2\ge 0$,

$g(x)\ge 1,$

with equality at $x=-1$. So the minimum value is

$m=3^1=3,$

and the minimum occurs at

$a=-1.$

Step 2: Find the maximum on $[-2,2]$

The quadratic opens upward, so the maximum on a closed interval occurs at an endpoint.

Compute:

$g(-2)=(-2)^2+2(-2)+2=2,$

$g(2)=2^2+2(2)+2=10.$

So the maximum occurs at $x=2$, hence

\qquad M=3^{10}.$$ Therefore, $$m+b=3+2=5.$$ **Answer: $5$**. ### Pitfall alert Do not take the derivative of $3^{x^2+2x+2}$ and then overthink it. Here the exponent is a simple quadratic, so completing the square is faster and cleaner. Also, remember that for a base greater than 1, the exponential keeps the same order as its exponent. ### Try different conditions If the base were $0<k<1$ instead of $3$, then the function would reverse the ordering of the exponent: the smallest exponent would give the largest function value. If the interval were open, the maximum might not exist at all, while the minimum would still occur at $x=-1$ because it lies inside the interval. ### Further reading completing the square, endpoint test, monotonic exponential