1.

If f (x) = [4 – (x – 7)3]1/5 is a real valued function, then show that f is invertible and find f – 1.

Answer»

For f to be invertible, f should be bijective. 

One-One 

Let x1, x2 be any arbitrary value ∈ R, then 

f (x1) = f (x2) 

⇒ [4 – (x1 – 7)3]1/5 = [4 – (x2 – 7)3]1/5 

⇒ 4 – (x1 – 7)3 = 4 – (x2 – 7)3 (Putting both the sides to power 5) 

⇒ (x1 – 7)3 = (x2 – 7)3 

⇒ x1 – 7 = x2 – 7 (Taking cube root of both the sides) 

⇒ x1 = x2 

⇒ f is one-one.

Onto 

Let y = f (x) = [4 – (x – 7)3]1/5 

⇒ y5 = 4 – (x – 7)3 ⇒ (x – 7)3 = 4 – y5

⇒ x - 7 = \(\sqrt[3]{4-y^5} \implies x = \sqrt[3]{4-y^5}\) + 7 ... (i)

Thus for every y ∈ R, there exists an x = \(\big(\sqrt[3]{4-y^5} +7\big)\) ∈ R 

⇒ f is onto.

∴  From (i)  x = f -1(y) = \(\big(\sqrt[3]{4-y^5} +7\big)\) 

or, f -1(x) = \(\big(\sqrt[3]{4-y^5} +7\big)\) (substituting x for y).


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