I must be losing my mind because I've been over and over this one and can't see what I am doing wrong:

Find the linearization L(x) of the function $\displaystyle f(x) = cos(x)$ at $\displaystyle a = \frac{3\pi}{2}$

Here is what I have done:

$\displaystyle f'(x) = -sin(x)$

$\displaystyle f'(\frac{3\pi}{2}) = -(-1) = 1$

$\displaystyle L(x) = f(x) + f'(x)(x-a)$

$\displaystyle L(x) = cos(\frac{3\pi}{2}) + 1(x - \frac{3\pi}{2})$

$\displaystyle L(x) = 0 + x - \frac{3\pi}{2}$

$\displaystyle L(x) = x - \frac{3\pi}{2}$

And that's my answer. But when I plug it into the web site, it gets kicked back as incorrect. Can anybody see my mistake? I can't!

Thanks.