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PolarNik [594]
2 years ago
5

Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.

Mathematics
1 answer:
masya89 [10]2 years ago
5 0

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is g^{-1}(x) = e^{x}

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

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)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is g^{-1}(x) = e^{x}

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y =  ln(x)

Taking exponents of both sides, we have

e^{y} = e^{lnx} \\e^{y} = x

Replacing x with y, we have

y = e^{x}

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is g^{-1}(x) = e^{x}

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(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 <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

brainly.com/question/9028678

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