Equation for tangent line of inverse function.

**calculuskangaroo** · September 15th 2011, 11:12 AM

Let f(x)=(x^4)+(x^3)+1
Let g(x) be the inverse of f(x) and define F(x)=f(2g(x)). Find an equation for the tangent line to y=F(x) at x=3.

The answer the book gives is y=(88x-89)/7 I just have no clue how to get there! Please help!

**HallsofIvy** · September 15th 2011, 11:25 AM

If $y= f(2f^{-1}(x))$ , by the chain rule, the derivative is $y'= f'(2f^{-1}(x))(2(f^{-1}(x)'$ . Further, ( $(f^{-1})'= \frac{1}{f'(y)}$ where y is such that f(y)= x. Here, $f(x)= x^4+ x^3+ 1$ so $f'(x)= 4x^3+ 3x^2$ . g(x) will be the value, y, such that f(y)= $y^4+ y^3+ 1= 3$ . Fortunately, it is clear that y= 1 satisfies that equation so $g(3)= f^{-1}(3)= 1$

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