Question

Find the critical points and stationary point of the function $f(x) = \dfrac{{{e^x}}}{x}$.

Answer 1

Here, we have given the function $f(x) = \dfrac{{{e^x}}}{x}$ and we have to find the critical point and the stationary point. critical points will be found by taking the derivative of the function equal to zero. The derivative of the division function is given by $d\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x) \times df(x) - f(x) \times dg(x)}}{{{{(g(x))}^2}}}$

Complete step by step solution:
Here, $f(x) = \dfrac{{{e^x}}}{x}$
Now, finding the derivative of the function,
$\Rightarrow {f^{'}}(x) = \dfrac{{{{({e^x})}^{'}}(x) - ({e^x}){{(x)}^{'}}}}{{{x^2}}} \\\ \Rightarrow {f^{'}}(x) = \dfrac{{{e^x}(x) - {e^x}}}{{{x^2}}} \\\ \Rightarrow {f^{'}}(x) = \dfrac{{{e^x}(x - 1)}}{{{x^2}}} \\\$
To find critical point taking ${f^{'}}(x) = 0$

$$\Rightarrow {f^{'}}(x) = 0 \\\ \Rightarrow \dfrac{{{e^{x}}(x - 1)}}{{{x^2}}} = 0 \\\ \Rightarrow {e^{x}}(x - 1) = 0 \\\$$

$\Rightarrow {e^{x}} = 0$
$\Rightarrow x = 1$
Now, $\Rightarrow {e^{x}} = 0$
Applying log on both sides,
$\Rightarrow \log {e^x} = \log 0 \\\ \Rightarrow x\log e = 0 \\\ \Rightarrow x = 0 \\\$

Note:
Critical point means where the derivative of the function is either zero or nonzero, while the stationary point means the derivative of the function is zero only. In the derivative of the division first of all we will derive the numerator and multiply by the denominator minus derivative of the denominator into numerator and divide whole by the square of the denominator.