Question

Using the Intermediate Value Theorem and a calculator, find an interval of length 0.01 that contains a root of e^x = 2 - x, rounding interval endpoints off to the nearest hundredth.

Answer 1

The function $g(x)=e^x$ is strictly increasing. The function $h(x)=2-x$ is strictlydecreasing.

Then if the root of $e^x=2-x$ exists, it will be the only root.

The function$f(x)=e^x-(2-x)$ is continuous on $(-\infin, \infin).$

$$f(0)=e^{0}-(2-0)=-1<0$$

$$f(0.5)=e^{0.5}-(2-0.5)=e^{0.5}-1.5=\sqrt{e}-1.5>0$$

$$f(0.2)=e^{0.2}-(2-0.2)=e^{0.2}-1.8\approx-0.5789<0$$

$$f(0.4)=e^{0.4}-(2-0.4)=e^{0.4}-1.6\approx-0.108<0$$

By the Intermediate Value Theorem exists a number $c\in (0.4,0.5)$ such that $f(c)=0$

$$f(0.45)=e^{0.45}-(2-0.45)=e^{0.45}-1.55\approx0.018>0$$

$$f(0.42)=e^{0.42}-(2-0.42)=e^{0.42}-1.58\approx-0.058<0$$

$$f(0.44)=e^{0.44}-(2-0.44)=e^{0.44}-1.56\approx-0.007<0$$

By the Intermediate Value Theorem exists a number $c\in (0.44,0.45),$ such that $f(c)=0.$

$(0.44,0.45)$