Question

How do you find the linearization at a=pi/4 of `f(x) = cos^(2)(x)`?

Answer 1

the linearization of f(x) is `y=-x+pi/4+1/2`

Explanation:

the solution:

Given: `f(x)=cos^2 x` at `a=pi/4`

`f(a)=cos^2 a`
`f(pi/4)=cos^2 (pi/4)=(1/sqrt2)^2=1/2`

We now have the point `(a, f(a))=(pi/4, 1/2)`

Compute for the slope `f'(a)`

`f' (x)=2*(cos x)(-sin x)`

`f' (a)=2*(cos a)(-sin a)`

`f' (pi/4)=2*(cos (pi/4))(-sin (pi/4))`

`f' (pi/4)=2*(1/sqrt2)(-1/sqrt2)=-1`

the linearization of `f(x)`

`y-f(a)=f' (a)(x-a)`

`y-1/2=-1(x-pi/4)`

`y=1/2-x+pi/4`

`y=-x+pi/4+1/2`

God bless....I hope the explanation is useful.