Proving the presence of one real root "Show that the following equation has exactly one real root:
$3x+2\cos x+5=0$"
So I think the way to approach this problem is to use a combination of the intermediate value theorem and the mean value theorem.
But since we are not given a specific interval, I'm not entirely sure what values of $f(a)$ and $f(b)$ to choose. Am I suppose to choose $f(0)$ and $f(1)$ and see that $0$ lies in between them. If so, how exactly can we proceed with the mean value theorem after using the intermediate value theorem. Do we have to check for continuity and the number $c$, or is there some other way?
Any help?
 A: Let $f(x)=3x+2\cos x+5$. Then $f'(x)=3-2\sin x>0$. Thus $f$ is strictly increasing. Then graph of $f$ can intersect the $x$-axis at most once.
Note $f$ is continuous and $\lim_{x\rightarrow-\infty}f(x)=-\infty$ and $f(0)=5>0$, hence it's graph must intersect the $x$-axis at least once.
A: Guide:
Let $f(x)=3x+2\cos x + 5$,
$$f'(x) = 3-2\sin(x)>0$$
Also compute limit of $x$ when $x$ tends to $-\infty$ and also $\infty$.
A: Hint:
Let $f(x) = 3x + 2\cos x + 5$. Note that $f(−π) < 0$ and $f(0) > 0$, which shows that there exists $c \in [−π,0]$ such that $f(c) = 0$, by the Intermediate Value Theorem. So $f(x)$ has at least one solution. 
Suppose that there are two distinct solutions $c_1 < c_2$ of this equation. What can you now conclude from Rolle’s Theorem?
A: The intermediate value theorem demonstrates one root: 
For example $f(-10)= -30 +2 \cos(-10) +5 \le 23 \lt 0$ while $f(10)= 30 +2 \cos(10) +5 \ge 33 \gt 0$ 
so there is a an intermediate value in $[-10,10]$ with $f(x)=0$ 

If using the mean value theorem then suppose there were two distinct roots $x_1$ and $x_2$ in that order.  Note $f(x)$ is continuous and differentiable for all $x$ 
You would then have some value $x_m \in [x_1,x_2]$ with $f'(x_m)= \dfrac{f(x_2)-f(x_1)}{x_2-x_1}=0$
But actually differentiating, $f'(x_m)= 3-2\sin(x_m) \ge 3-2=1 \gt 0$, leading to the necessary contradiction

Thus there is at least one root and there are not two distinct roots, so there is exactly one
A: $2 \cos x$ is between -2 and 2 inclusive for all real $x$.
Therefore 
If $3x+5>2$ then the result is positive.
If $3x+5<-2$ then the result is negative.
Which simplifies to 
If $x > -1$ the result is positive.
If $x < -2\frac{1}{3}$ the result is negative.
This tells us two things.


*

*By the intermediate value theorem there must be at least one root in the range $-2\frac{1}{3}$ to $-1$

*All roots must be in the range $-2\frac{1}{3}$ to $-1$


We now observe that $3x+5$ is monotonically increasing for all x and that  $\cos x$ is monotonically increasing from $-\pi$ to $0$. Therefore our function is monotonically increasing over the range $-2\frac{1}{3}$ to $-1$. Therefore there cannot be more than one root in the range $-2\frac{1}{3}$ to $-1$ therefore there cannot be more than one root.
We have observed that there must be at least one root and that there cannot be more than one root. Therefore there must be exactly one root.
A: A good way to see there is exactly one real root is to sketch the cosine curve $y=2\cos x$ and the straight line $y=-3x-5$ and see that they cross only once, somewhere between $-\pi$ and $0$.
To make this somewhat more rigorous, note that $|2\cos x|\le2$ for all $x$ and $|3x+5|\gt2$ for $x$ outside of $[-\pi,0]$ (actually outside of $[-{7\over3},-1]$, but $[-\pi,0]$ is good enough), so they can only intersect in the interval $[-\pi,0]$.  The cosine function $y=2\cos x$ is strictly increasing on $[-\pi,0]$, while the straight line $y=-3x-5$ is strictly decreasing for all $x$, so the two curves cannot cross more than once.  Finally, $y=2\cos x$ goes from negative to positive while $y=-3x-5$ goes from negative to positive, so they must cross somewhere.  But all this is more or less obvious if you just sketch the two graphs.
