# Question 96b06

Oct 19, 2016

See below.

#### Explanation:

${e}^{- x}$ is monotonic strictly decreasing for $x \in \left[0 , \infty\right)$ and
${x}^{3}$ is monotonic strictly increasing for $x \in \left[0 , 1\right)$ both functions are continuous so they must cross for a point ${x}_{0}$ such that
${e}^{- {x}_{0}} = {x}_{0}^{3} < 1$. Note that for $x \in \left[0 , 1\right)$ we have
${x}_{0}^{3} < 1$ and ${e}^{- {x}_{0}} < 1$

Oct 19, 2016

#### Explanation:

You must use the intermediate value theorem
let $f \left(x\right) = {e}^{-} x - {x}^{3}$ this is a continuous function on the interval $\left(0 , 1\right)$
$f \left(0\right) = {e}^{0} - 0 = 1$
and $f \left(1\right) = {e}^{-} 1 - 1 < 0$
Then there is a value c ∈ (0,1) # such that $f \left(1\right) < f \left(c\right) < f \left(0\right)$

Oct 26, 2016

It is explained below

#### Explanation:

Given ${e}^{- x} = {x}^{3}$, take natural log on both sides. It would be $- x \ln e = 3 \ln x$
Or -x = 3 ln x. This expression signifies that for all real x, x cannot be less than or equal to 0.

Now, in the logarithmic equation the LHS has - x, while it has a positive expression 3ln x on the RHS. This would be possible only if x is less than 1

Thus it is proved that x would lie between 0 and 1