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