Question

find the exact location of all the relative and absolute extrema of the function $f(x)=e^{x}-2.3x$ with domain $-1,1$.
$f$ has an

at $(x,y)=$(, (smallest $x$-value)
$f$ has an

at $(x,y)=$(, (largest $x$-value)

Explanation:

Step1: Find the derivative

The derivative of $f(x)=e^{x}-2.3x$ is $f^\prime(x)=e^{x}-2.3$.

Step2: Find critical points

Set $f^\prime(x) = 0$, so $e^{x}-2.3 = 0$. Then $e^{x}=2.3$, and $x=\ln(2.3)\approx0.8329$. Since $\ln(2.3)\in[- 1,1]$, it is a critical - point.

Step3: Evaluate the function at critical and end - points

Evaluate $f(x)$ at $x=-1$, $x = \ln(2.3)$, and $x = 1$.
$f(-1)=e^{-1}+2.3=\frac{1}{e}+2.3\approx0.3679 + 2.3=2.6679$.
$f(\ln(2.3))=e^{\ln(2.3)}-2.3\ln(2.3)=2.3-2.3\ln(2.3)\approx2.3-2.3\times0.8329=2.3 - 1.9157=0.3843$.
$f(1)=e - 2.3\approx2.7183-2.3 = 0.4183$.

Answer:

$f$ has an absolute minimum at $(x,y)=(\ln(2.3),2.3 - 2.3\ln(2.3))$ (smallest $x$-value among relevant extrema in the domain).
$f$ has an absolute maximum at $(x,y)=(1,e - 2.3)$ (largest $x$-value among relevant extrema in the domain).