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Show the the following equation has a real solution $$\sin(x^2)=x^3+2x^2+1.$$

I understand I need to use intermediate value theorem to show it has a real solution, but I'm not sure how to proceed as the function is neither increasing or decreasing.

What I have done: let $f(x)=x^3+2x^2+1-\sin(x^2)$ then $f'(x)=3x^2+4x-2x\cos(x^2)$ which cannot be determined as either increasing or decreasing.

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We show that $f(x)=x^3+2x^2+1-\sin(x^2)$ has a unique zero in $\mathbb{R}$.

Since $f(-3)<0$ and $f(-2)>0$, the existence of at least a real solution in $(-3,-2)$, follows from the intermediate value theorem.

Moreover, $f(0)=1$ and for $x\in (-2,0)\cup (0,+\infty)$, $$f(x)=\underbrace{x^2(x+2)}_{>0}+\underbrace{(1-\sin(x^2))}_{\geq 0}>0.$$

As you already noted $f'(x)=3x^2+4x-2x\cos(x^2)$. Then, for $x<-2$, $$f'(x)=\underbrace{x}_{<0}(\underbrace{3x+2}_{< -4}+\underbrace{2(1-\cos(x^2))}_{\leq 4})>0$$
which implies that $f$ is strictly increasing in $(-\infty,-2]$ and it has a unique zero.

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Hint:

Compute $f(0)$ and consider $\lim_{x\to-\infty} f(x)$.

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Actually it has nothing to do with derivatives.

Let $f(x)=x^3+2x^2+1-\sin(x^2)$. Easy to notice that $$f(0)>0 \ , \ f(-3)<0$$ hence $f(x)$ vanishes for some real $x\in (-3,0)$.

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Hint: $f(0)>0$ and $f(-3)<0$.

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Hint

Observe that $$f(0)\gt 0$$ and $$f(-\pi) \lt 0$$

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