Here's the question you clicked on:
rmalik2
Find the linearization L(x) of the function at a = 7π/2 f(x)=cos(x)
We need a point and a slope to find a linearization. All you are really finding is a y=mx+b First find m Take the derivative of cos x sub in 7pi/2 This value will be the slope for your line Now we need a point We know x = 7pi/2 Use f(x)=cos(x) to find y = f(7pi/2) Then I like to use the point-slope form of the equation of a line \[y-y _{1} =m \left( x-x _{1} \right) \] Solve for y
can you show me steps by steps please?
\[f(x)=f(a)+f'(a)(x-a)=\cos{\frac{7\pi}{2}}-\sin{\frac{7\pi}{2}}\left(x-\frac{7\pi}{2}\right)=x-\frac{7\pi}{2}\]
derivative of cos(x) = -sin(x) -sin(7pi/2) = -sin(pi/2) = -(-1) now for the point (7pi/2, ? ) cos(7pi/2) = 0 Then it is just like what nikvist posted. OK?