a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
b. The inverse image of g(x) is [tex]g^{-1}(x) = e^{x}[/tex]
c. The inverse image of h(x) is h⁻¹(x) = x + 9
The domain of T
Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0
⇒ x² ≥ 25
⇒ x ≥ ±5
⇒ -5 ≤ x ≤ 5.
Inverse image of f(x)
The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
f : R → R defined by f(x) = 3x³
Let f(x) = y.
So, y = 3x³
Dividing through by 3, we have
y/3 = x³
Taking cube root of both sides, we have
x = ∛(y/3)
Replacing y with x we have
y = ∛(x/3)
Replacing y with f⁻¹(x), we have
So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)
Inverse image of g(x)
The inverse image of g(x) is [tex]g^{-1}(x) = e^{x}[/tex]
g : R+ → R defined by g(x) = ln(x).
Let g(x) = y
y = ln(x)
Taking exponents of both sides, we have
[tex]e^{y} = e^{lnx} \\e^{y} = x[/tex]
Replacing x with y, we have
[tex]y = e^{x}[/tex]
Replacing y with g⁻¹(x), we have
So, the inverse image of g(x) is [tex]g^{-1}(x) = e^{x}[/tex]
Inverse image of h(x)
The inverse image of h(x) is h⁻¹(x) = x + 9
h : R → R defined by h(x) = x − 9
Let y = h(x)
y = x - 9
Adding 9 to both sides, we have
y + 9 = x
Replacing x with y, we have
x + 9 = y
Replacing y with h⁻¹(x), we have
So, the inverse image of h(x) is h⁻¹(x) = x + 9
Learn more about inverse image of a function here:
https://brainly.com/question/9028678
