Find the roots of 2 equations

Question

Show that the equation $e^{-x} = x^2$ has a root between $x=0.70$ and $x=0.71$.

I think you have to use natural logs to get rid of the $e$ however after that, i'm not sure how to solve for $x$

Answer 1

Hint: $f(x)=e^{-x}-x^2$ is continuous, and $f(0.70)>0>f(0.71),$ so....

Answer 2

Here is one way. Let $f(x) = e^{-x} - x^2$.

Note that $f(0.70) >0$ and $f(0.71) <0$. Hence since the conditions of the intermediate value theorem apply, we must have $x \in (0.70,0.71)$ such that $f(x) = 0$.

If you prefer, you could take logs to get the equation $-x = 2 \ln x$. Then let $g(x) = x+2 \ln x$ and note that $g(0.70)<0$ and $g(0.71)>0$.

Answer 3

I have found how to solve this question. Since the equation has a root, means it crosses the x axis and goes to either the + or - side of the y axis. Therefore to show it has a root between 0.71 and 0.70, we substitute each of the values into the equation and show that the answers has different signs.