Asked by Rayan
Suppose f(x)= x^5 - 10 and let h(x) be the inverse of f. Find h'(22).
Answers
Answered by
Reiny
f(x) = x^5 - 10
inverse is x = y^5 - 10
y^5 = x+10
y = (x+10)^(1/5)
h(x) = (x+10)^(1/5)
h'(x) = (1/5)(x+10^(-4/5)
h'(22) = (1/5)(32)^(-4/5)
= (1/5)(2)^-4
= (1/5)(1/16)
= 1/80
inverse is x = y^5 - 10
y^5 = x+10
y = (x+10)^(1/5)
h(x) = (x+10)^(1/5)
h'(x) = (1/5)(x+10^(-4/5)
h'(22) = (1/5)(32)^(-4/5)
= (1/5)(2)^-4
= (1/5)(1/16)
= 1/80
Answered by
Previous Physics Student
Thank you that helps a lot!
Answered by
Steve
If f(a) = b then h'(b) = 1/f'(a)
f(2) = 22
f'(2) = 80
h'(22) = 1/f'(2) = 1/80
f(2) = 22
f'(2) = 80
h'(22) = 1/f'(2) = 1/80
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