calculus
Sketch the graph of $$f(x)=1+\frac ax+\frac {a} {x^2}$$, $a \gt0$
how is it even possible to draw a graph with constant? I mean how can I sketch it? it with different values of a my graph and its asymptotes will change, at least the horizontal one.
ok, here I can say for sure that vertical asymptote is $x=0$. But how to proceed with critical and inflection points? how can I tag them on graph?
For any $x\in [0,2\pi]$ you get $f'(x)=\cos (x) -\sin (x)$ and $f''(x)=-\sin (x)-\cos (x)$. Therefore $$\begin{align} f'(x)=0&\iff \cos (x) -\sin(x)=0\\ &\iff \cos (x) =\sin (x)\\ &\iff x\in \{\frac{\pi}{4}, \frac{5\pi}{4}\} \end{align}$$
Which gives you the critical points $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{5\pi}{4}$.
The last equivalence is easy to see geometrically if you look at the unit circle.
I don't understand why you're messing with $\cos ^{-1}$. Care to explain so we can help you?
Similarly, for $f''$ you'll get the zeros $\displaystyle \frac{3\pi}{4}$ and $\displaystyle \frac{7\pi}{4}$.
This should bring you forward:
However, you should - at least once - really understand those criteria (e.g. graphically) rather than just apply them.
Best Answer
Setting $f(x) = 0$, $$\begin{align*} 1+\frac ax + \frac a{x^2} &= 0\\ x^2+ax+a &= 0\\ x &= \frac{-a\pm\sqrt{a^2-4a}}{2} \end{align*}$$ So the $x$-intercepts, if any, depend on $a$.
Differentiating $f(x)$ w.r.t. $x$, $$f'(x) = -\frac a{x^2}-\frac {2a}{x^3} = -\frac{ax(x+2)}{x^4}$$ Setting $f'(x) = 0$, $$\begin{align*} -\frac a{x^2}-\frac {2a}{x^3} &= 0\\ -ax-2a&= 0\\ x&=-2 \end{align*}$$ The $x$-coordinate of the stationary point does not depend on $a$, but the $y$-coordinate does.
Differentiating $f'(x)$ w.r.t. $x$, $$f''(x) = \frac{2a}{x^3} + \frac{6a}{x^4} = \frac{2a(x+3)}{x^4}$$ Setting $f''(x) = 0$, $$\begin{align*} \frac{2a}{x^3} + \frac{6a}{x^4} &= 0\\ 2ax + 6a &= 0\\ x&= -3 \end{align*}$$ The $x$-coordinate of the inflexion point does not depend on $a$, but the $y$-coordinate does.
Consider the signs in terms of $x$, $$\begin{array}{r|c|c|c|c|c|c} x&(-\infty,-3)&-3&(-3,-2)&-2&(-2,0)&(0,\infty)\\\hline f(x)&&1-\frac a3+\frac a9&&1-\frac a2+\frac a4&&+\\\hline f'(x)&-&-&-&0&+&-\\\hline f''(x)&-&0&+&+&+&+ \end{array}$$
And also the horizontal asymptote. $$\lim_{x\to \infty}\left(1+\frac ax+\frac a{x^2}\right) = \lim_{x\to\infty} 1 + \lim_{x\to\infty} \frac ax+ \lim_{x\to\infty}\frac a{x^2} = 1$$