Let f be the function defined by f(x)=x^3+x^2+x. Let g(x)=f^-1(x), where g(3)=1. What is the value of g'(3)?

1/6

Step-by-step explanation:

Given the function

f(x)=x^3+x^2+x

Get the inverse. Let y = f(x)

y = x^3+x^2+x

Replace y with x

x = y³+y²+y

Differentiate with respect to x

1 =3y²dy/dx + 2ydy/dx + dy/dx

1 = (3y²+2y+1)dy/dx

dy/dx = 1/3y²+2y+1

g'(x) = 1/3(g(x))²+2g(x)+1

If x = 3

g'(3) = 1/3(g(3))²+2g(3)+1

g'(3) =1/3(1)²+2(1)+1

g'(3) = 1/3+2+1

g'(3) = 1/6

Hence g'(3) = 1/6

The statement "y varies directly as x," means that when x increases, y increases by the same factor. In other words, y and x always have the same ratio: y/x =K

it can be written as ---> y=kx

then put in the values to get a common ratio
k= y/x
k1= 11/7
k2= 13/8

therefore, there's no common ratio in the given table