Question

if f(x)=x^3 + 1/x and g(x)=f^-1 (x) find the slope of the curve g(x) = f^-1(x) at the point (2,1).

Answer 1

\[Is~f(x) = \frac{ x^3+1 }{ x }~~or~~f(x) = x^3 + \frac{ 1 }{ x }~~?\]

Answer 2

I didn't even see the \(/x\).

Answer 3

\[x^3 + \frac{ 1 }{ x }\]

Answer 4

They want you to find g'(x) at the point (2,1).
If (2,1) is a point on the inverse function g(x),
then (1,2) is a point on the original function f(x).
g'(x) at (2,1) = 1 / { f'(x) at (1,2) }

Answer 5

d/dx (f^-1(a)) = 1/ f' (f^-1(a) )


g(2) = 1 implies f(1) = 2

so d/dx (f^-1(1) ) = 1/ f' (f^-1(1) ) = 1 / f'(2)

f'(x) = 3x^2 - 1/x^2
f'(2) = 3(2)^2 - 1/(2)^2 = 47/4

d/dx (f^-1(1)) = 1 / (47/4)) = 4/47

Answer 6

y u do dis sourwing

Answer 7

what do you mean?

Answer 8

options are 1, 1/2, 1/3 , -1/2

Answer 9

oops i evaluate at the wrong point
d/dx (f^-1(2)) = 1 / f' (f'^-1)(2) = 1/ f'(1)

f'(1) = 2

so 1/2