# Is f(x)= cos(x+(5pi)/4)  increasing or decreasing at x=-pi/4 ?

Jul 9, 2018

See explanation.

#### Explanation:

Generally if the function $f \left(x\right)$ has the derrivative ${f}^{'} \left({x}_{0}\right)$ then we can say that:

• $f \left(x\right)$ is increasing at ${x}_{0}$ if ${f}^{'} \left({x}_{0}\right) > 0$

• $f \left(x\right)$ is decreasing at ${x}_{0}$ if ${f}^{'} \left({x}_{0}\right) < 0$

• ${f}^{'} \left(x\right)$ may have an extremum at ${x}_{0}$ if ${f}^{'} \left({x}_{0}\right) = 0$ (additional test is required)

In the given example we have:

${f}^{'} \left(x\right) = - \sin \left(x + \frac{5 \pi}{4}\right)$

${f}^{'} \left({x}_{0}\right) = - \sin \left(- \frac{\pi}{4} + \frac{5 \pi}{4}\right) = - \sin \left(\pi\right) = 0$

${f}^{'} \left({x}_{0}\right) = 0$, so $f \left(x\right)$ has either an extremum, or an inflection point.
To check if the point is extremum we have to check if the first derivative changes sign at ${x}_{0}$.

graph{(y+sin(x+(5pi)/4))((x+pi/4)^2+(y^2)-0.01)=0 [-4, 4, -2, 2]}

At ${x}_{0} = - \frac{\pi}{4}$ the derrivative changes sign from negative to positive, so the point is a minimum.