Question

What is the inverse of the function f(x) = 2x – 10? h(x) = 2x – 5 h(x) = 2x + 5 h(x) = one-halfx – 5 h(x) = one-halfx + 5

Answer 1

Answer:

$\text{Inverse of f(x) is }f^{-1}(x)=\frac{x}{2}+5$

Step-by-step explanation:

$\text{Given}$

$f(x)=2x-10=y(say)$

$\text{clearly, f(x) is one-one and onto}$

$\implies\:2x-10=y(say)$

$\implies\:2x=y+10(say)$

$\implies\:x=\frac{y+10}{2}(say)$

$\implies\:x=\frac{y}{2}+5(say)$

$\text{Hence, inverse of f(x) is }f^{-1}(x)=\frac{x}{2}+5$

Answer 2

Solution:

The given function is f(x) = 2x - 10

First we show that f is invertible.

1) Let x₁, x₂ be two distinct elements in the set of real numbers, taken as the domain of f.

f(x₁) = 2x₁ - 10, f(x₂) = 2x₂ - 10

f(x₁) - f(x₂) = 2 (x₁ - x₂) ≠ 0 [∵ x₁ ≠ x₂]

Since x₁ ≠ x₂ gives f(x₁) ≠ f(x₂), f is injective.

2) Let y be an arbitrary element in the set of real numbers, taken as the co-domain of f.

f(x) = y

or, 2x - 10 = y

or, x = (y + 10)/2

Since y is a real number, (y + 10)/2 is also a real number. Therefore y has a pre-image (y + 10)/2 in the domain of f. Since y is taken as arbitrary, each element in the co-domain of f has a pre-image under f. Therefore f is surjective.

Since f is injective and surjective, f is a bijection, and hence invertible.

We have found that each element y in the co-domain of f has a unique pr-image (y + 10)/2

So f⁻¹ is defined by

f⁻¹(y) = (y + 10)/2, y is a real number

or, equivalently f⁻¹(x) = 1/2 x + 5

Therefore, h(x) = (1/2) x + 5,

which is the required inverse function.