find the exact location of all the relative and absolute extrema of the function f(x)=e^x - 1.9x with domain…

Explanation:

Step1: Find the derivative

The derivative of $f(x)=e^{x}-1.9x$ is $f^\prime(x)=e^{x}-1.9$.

Step2: Set the derivative equal to zero

Set $e^{x}-1.9 = 0$. Then $e^{x}=1.9$, and solving for $x$ gives $x=\ln(1.9)$.

Step3: Evaluate the function at critical - point and endpoints

Evaluate $f(x)$ at $x = - 1,x=\ln(1.9),x = 1$.

• When $x=-1$, $f(-1)=e^{-1}+1.9=\frac{1}{e}+1.9\approx0.368 + 1.9=2.268$.
• When $x=\ln(1.9)$, $f(\ln(1.9))=1.9-1.9\ln(1.9)=1.9(1 - \ln(1.9))\approx1.9(1 - 0.642)=1.9\times0.358 = 0.6802$.
• When $x = 1$, $f(1)=e - 1.9\approx2.718-1.9 = 0.818$.

Step4: Determine the extrema

The smallest $x$ - value for an extreme is $x=-1$ and $f(-1)\approx2.268$, so the point is $(-1,e^{-1}+1.9)$. The largest $x$ - value for an extreme is $x = 1$ and $f(1)=e - 1.9$, so the point is $(1,e - 1.9)$.

Answer:

Smallest $x$ - value: $(-1,e^{-1}+1.9)$ Largest $x$ - value: $(1,e - 1.9)$