How many real solutions does this equation have: $x^{2}-2\sin x+2x=33$
I have created this equation myself, so it's not homework and I hope it's even solvable.. It will be only for practice.
$$x^{2}-2\sin x+2x=33$$
$$x^{2}-2\sin x+2x-33 = 0$$
$$f'(x) = 0$$
$$2x-2\cos x+2 =0$$
$$x = 0$$
Now check if it's maximum or minimum:
$$f''(x) = 2+2\sin x$$
$$f''(0) = 2+2\sin (0) > 0 \Rightarrow minimum$$
Now let's calculate the point of minimum:
$$0^{2}-2\sin (0)+2\cdot 0 -33= -33$$
$\Rightarrow P(0|-33)$
Now let's check if function is monotonic:
$f'(x) = 2x -2\cos x + 2$
For $x>0$ we get $f'(x) > 0$ , so monotonic increasing
For $x<0$ we get $f'(x) < 0$ , so monotonic decreasing
$\Rightarrow$ Function has $2$ real solutions
My questions:
Did I do it correctly?
If correct, is there something I could have skipped because not very necessary (I'm thinking to skip the calculation of maximum / minimum and just do the monotony)?
You're implicitly defining $f(x)=x^2-2\sin x+2x-33$; the derivative is $$ f'(x)=2x-2\cos x+2=2(1+x-\cos x) $$ and the second derivative is $$ f''(x)=1+\sin x $$ The second derivative is positive except at isolated points, so the function is convex and it can have at most a point of minimum. Since $$ \lim_{x\to-\infty}f(x)=\infty=\lim_{x\to\infty}f(x) $$ we know that the function has an absolute minimum. As $f'(0)=0$, the minimum is located at $0$. Now $$ f(0)=-33<0 $$ so the equation $f(x)=0$ has two solutions.
The problem with your solution is that you didn't prove the only solution of $f'(x)=0$ is $x=0$. This follows from convexity, which is the same as saying that $f'$ is an increasing function. You state, without proof, that $f'(x)<0$ for $x<0$, which is easy, but not trivial.