What are the absolute extrema of f(x)=x - e^x in[1,ln8]?

Mar 24, 2016

There is an absolute maximum of $- 1.718$ at $x = 1$ and an absolute minimum of $- 5.921$ at $x = \ln 8$.

Explanation:

To determine absolute extrema on an interval, we must find the critical values of the function that lie within the interval. Then, we must test both the endpoints of the interval and the critical values. These are the spots where critical values could occur.

Finding critical values:

The critical values of $f \left(x\right)$ occur whenever $f ' \left(x\right) = 0$. Thus, we must find the derivative of $f \left(x\right)$.

If:$\text{ "" "" "" "" } f \left(x\right) = x - {e}^{x}$
Then: $\text{ "" "" } f ' \left(x\right) = 1 - {e}^{x}$

So, the critical values will occur when: $\text{ "" } 1 - {e}^{x} = 0$
Which implies that:$\text{ "" "" "" "" "" "" "" "" } {e}^{x} = 1$
So:$\text{ "" "" "" "" "" "" "" "" "" "" "" "" "" } x = \ln 1 = 0$

The function's only critical value is at $x = 0$, which is not on the given interval $\left[1 , \ln 8\right]$. Thus, the only values at which the absolute extrema could occur are at $x = 1$ and $x = \ln 8$.

Testing possible values:

Simply, find $f \left(1\right)$ and $f \left(\ln 8\right)$. The smaller is the function's absolute minimum and the larger is the absolute maximum.

$f \left(1\right) = 1 - {e}^{1} = 1 - e \approx - 1.718$

$f \left(\ln 8\right) = \ln 8 - {e}^{\ln} 8 = \ln 8 - 8 \approx - 5.921$

Thus, there is an absolute maximum of $- 1.718$ at $x = 1$ and an absolute minimum of $- 5.921$ at $x = \ln 8$.

Graphed is the original function on the given interval:

graph{x-e^x [.9, 2.079, -7, 1]}

Since there are no critical values, the function will remain decreasing over the entire interval. Since $x = 1$ is the beginning of the constantly decreasing interval, it will have the highest value. The same logic applies to $x = \ln 8$, since it is the furthest of the interval and will be the lowest.