Here, g(f(x))=x as f(x) and g(x) are inverse of each other.
Now, g′(f(x))f′(x)=1
⇒g′(f(x))=f′(x)1…(i)
Now f(x)=63⇒x3+x−5=63
⇒x3+x−68=0
So x=4 satisfies the above equation
g′(63)=f′(4)1 from (i)
=3(4)2+11=491
Let f:R→R be defined as f(x)=x3+x−5. If g(x) is a function such that f(g(x))=x,∀x∈R, then g′(63) is equal to ______
Held on 25 Jun 2022 · Verified 6 Jul 2026.
49
491
4943
493
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