Question

f (x) = - eˣ Baseline (x - 4)
What​ is(are) the critical​ point(s) of​ f?

Answer 1

Answer

given,

   $f(x) = \dfrac{-e^x}{x - 4}$

to find the critical point of the given expression

fist differentiating the function

$f'(x) = -\dfrac{(x-4)e^x+ e^x}{(x - 4)^2}$

$f'(x) = \dfrac{-(x-4)e^x- e^x}{(x - 4)^2}$

$f'(x) = \dfrac{-e^x(x-3)}{(x - 4)^2}$

now equating differential equation to zero

$\dfrac{e^x(-x+3)}{(x - 4)^2}=0$

$e^x(-x+3)=0$

now,

-x + 3 = 0            and eˣ ≠ 0

x = 3          

the critical number will be equal to x = 3

$y = \dfrac{-e^3}{3 - 4}$

$y =e^3$