Q. 94.0( 4 Votes )
Consider f : R+→ [–5, ∞) given by f(x) = 9x2 + 6x – 5. Show that f is invertible with.
We have f : R+→ [–5, ∞) and f(x) = 9x2 + 6x – 5.
Recall that a function is invertible only when it is both one-one and onto.
First, we will prove that f is one-one.
Let x1, x2ϵ R+ (domain) such that f(x1) = f(x2)
⇒ 9x12 + 6x1 – 5 = 9x22 + 6x2 – 5
⇒ 9x12 + 6x1 = 9x22 + 6x2
⇒ 9x12 – 9x22 + 6x1 – 6x2 = 0
⇒ 9(x12 – x22) + 6(x1 – x2) = 0
⇒ 9(x1 – x2)(x1 + x2) + 6(x1 – x2) = 0
⇒ (x1 – x2)[9(x1 + x2) + 6] = 0
⇒ x1 – x2 = 0 (as x1, x2ϵ R+)
∴ x1 = x2
So, we have f(x1) = f(x2) ⇒ x1 = x2.
Thus, function f is one-one.
Now, we will prove that f is onto.
Let y ϵ [–5, ∞) (co-domain) such that f(x) = y
⇒ 9x2 + 6x – 5 = y
Adding 6 to both sides, we get
9x2 + 6x – 5 + 6 = y + 6
⇒ 9x2 + 6x + 1 = y + 6
⇒ (3x + 1)2 = y + 6
Clearly, for every y ϵ [4, ∞), there exists x ϵ R+ (domain) such that f(x) = y and hence, function f is onto.
Thus, the function f has an inverse.
We have f(x) = y ⇒ x = f-1(y)
But, we found f(x) = y ⇒
Thus, f(x) is invertible and
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