Show that f(x) = e^(1/x), x ≠ 0 is a decreasing function for all x ≠ 0.

Question

Show that \(\text{f(x)} = e^\frac{1}{\text{x}}, x \neq 0\) is a decreasing function for all x ≠ 0.

Answer 1

Given:- Function f(x) = e^{\(\frac{1}{\text{x}}\)}

Theorem:- Let f be a differentiable real function defined on an open interval (a, b).

(i) If f’(x) > 0 for all x ∈ (a, b), then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all x ∈ (a, b), then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

Their ratio is also greater than 0

as by applying –ve sign change in comparision sign

⇒ f’(x) < 0

Hence, condition for f(x) to be decreasing

Thus f(x) is decreasing for all x ≠ 0