If f(Pi/8) = 0 and f '(x) = 3/2 + (cos (2x))^2, find the equation of the tangent line to the graph of this function at the point (Pi/8, 0).

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- Oct 6th 2009, 05:27 PMzweevuFind equation of tangent line
If f(Pi/8) = 0 and f '(x) = 3/2 + (cos (2x))^2, find the equation of the tangent line to the graph of this function at the point (Pi/8, 0).

- Oct 6th 2009, 06:42 PMartvandalay11

A tangent line has the following form for all functions

$\displaystyle y-y_0=m(x-x_0)$

Where $\displaystyle (x_0,y_0)$ is the point at which you want the line to be tangent and m= the derivative evalauted at the point $\displaystyle (x_0,y_0)$

So $\displaystyle f'(\frac{\pi}{8})=\frac{3}{2}+\cos^2(2\cdot\frac{\ pi}{8})$

$\displaystyle =\frac{3}{2}+\cos^2(\frac{\pi}{4})=\frac{3}{2}+\le ft ( \frac{\sqrt{2}}{2}\right )^2$

$\displaystyle =\frac{3}{2}+\frac{2}{4}=2$

I leave it to you to plug it into the formula I provided