Answer to Question #260213 in Calculus for Almas

Question

Plot a graph of cosx

Answer 1

$$f(x)=\cos x$$

Domain: $(-\infin, \infin)$

Range: $[-1, 1]$

There are no asymptotes.

$$f(-x)=\cos(-x)=\cos x=f(x), x\in \R$$

The function is even. The graph is symmetric with respect to the y-axis.

$$f(x+T)=f(x+2\pi)=f(x), x\in \R$$

The function is periodic. The period is $T=2\pi.$

$y$ - intersection: $x=0, f(0)=\cos(0)=1.$

Point $(0, 1)$

$x$ - intersection(s): $y=0, 0=\cos x=>x=\dfrac{\pi}{2}+\pi n, n\in \Z.$

Points $(\dfrac{\pi}{2}+\pi n, 0), n\in \Z$

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

Find the critical number(s)

$$f'(x)=0=>-\sin x=0=>x=\pi k, k\in \Z$$

If $-\pi+2\pi k<x<2\pi k, k\in \Z,$ then $f'(x)>0, f(x)$ increases.

If $2\pi k<x<\pi+2\pi k, k\in \Z,$ then $f'(x)<0, f(x)$ decreases.

$f(2\pi k)=\cos(2\pi k)=1,k\in \Z$

$f(\pi+2\pi k)=\cos(\pi+2\pi k)=-1,k\in \Z$

The function $f(x)=\cos x$ has the local maximum with value of $1$ at $x=2\pi k, k\in \Z.$

The function $f(x)=\cos x$ has the local minimum with value of $-1$ at $x=\pi+2\pi k, k\in \Z.$

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

Find the point(s) of inflection

$$f''(x)=0=>-\cos x=0=>x=\dfrac{\pi}{2}+\pi m, m\in \Z$$ $$f(\dfrac{\pi}{2}+\pi m)=\cos(\dfrac{\pi}{2}+\pi m)=0,m\in \Z$$

Points of inflection: $(\dfrac{\pi}{2}+\pi m, 0), m\in\Z$

If $-\dfrac{\pi}{2}+2\pi m<x<\dfrac{\pi}{2}+2\pi m, m\in \Z,$ then $f''(x)<0, f(x)$ is concave down.

If $\dfrac{\pi}{2}+2\pi m<x<\dfrac{3\pi}{2}+2\pi m, m\in \Z,$ then $f''(x)>0, f(x)$ is concave up.

Sketch the graph of $f(x)=\cos x$