df/dx = 6(2x+1)^2
so, g' = 1/f'
g'(1) = 1/(6*9) = 1/54
check:
g(x) = (∛x-1)/2
g'(x) = 1/(6(∛x^2))
g'(1) = 1/(6*9) = 1/54
Let f(x)=(2x+1)^3 and let g be the inverse function of f. Given f(0)=1, what is the value of g'(1)?
A. -2/27
B. 1/54
C. 1/27
D. 1/6
E. 6
2 answers
f(x)=(2x+1)^3
the inverse is (2y+1)^3=x
the derivative of this is 6(dy/dx)(2y+1)^2=1
you know that g(1)=0, because it's the inverse of f(x)
6(dy/dx)(2(0)+1)^2=1
6(dy/dx)=1
1/6=dy/dx
so the answer would be D
I know there's a shortcut but I forget what it is
the inverse is (2y+1)^3=x
the derivative of this is 6(dy/dx)(2y+1)^2=1
you know that g(1)=0, because it's the inverse of f(x)
6(dy/dx)(2(0)+1)^2=1
6(dy/dx)=1
1/6=dy/dx
so the answer would be D
I know there's a shortcut but I forget what it is
7 answers