To find the number of zeroes P1 - Given $f (x) = x^ {3} + ax^ {2} +6x -1$ has critical point at $x=-2$, then how many real solutions has $f (x) =0$?
MY Attempt regarding is that first i have found value of a which is 9\2 by setting -2 in derivative , thean i also see that f (2) is positive and f(-1) is negative so one root is there between them for sure .Also this can be seen by using corollory of IVT .But how can i be sure of other roots ? 
Given $f(x) = xsinx + cosx$ , i have to find number of zeroes .Any hints for this problem .Thanks
 A: You have found that $a=\frac 92$, so the derivative being $3x^2-9x+6$ it also cancels for $x=-1$. So, now compute the value of the function at the critical points $$f(-2)=-3$$ $$f(-1)=-\frac{7}{2}$$ So, at both minimum and maximum, the function values are negative. Then only one real root is possible and, since by inspection $f(0)=-1$ and $f(1)=\frac{21}{2}$, the root should be positive, definitely smaller than $0$.
For your second problem, you could rewrite $x \sin(x)+\cos(x)=0$ as $x=-\cot(x)$ and consider the intersection of the two curves. Function $\cot(x)$ being undefined for $x=k\pi$, you have an infinite number of solutions, closer and closer to $k \pi$ (just smaller than).
A: plugging $-2$ in $f'(x)$ we get $a=\frac{9}{2}$ and we have $f'(x)=3x^2+9x+6$ and $f''(x)=6x+9$ plugging $x=-2$ in the second derivative we get $f''(-2)=-3<0$ and thus we have a local maximum at $(-2;-3)$ further we have for $x=-1$ a local minimum thus for $x>-1$ the function is strictly monotomously increasing and we have exactly one zero point,
since $\lim_{x\to \infty}f(x)=\infty$
