Show that function $4x^5+x^3+2x+1=0$ has exactly one real root. I am struggling with problems like this, just trying to grasp the concept.
Using the Intermediate Value Theorem and Rolle's Theorem, prove that $4x^5+x^3+2x+1=0$ has exactly one real root.

Let $f(x) = 4x^{5} + x^{3} + 2x + 1$. As noted in other answer there is a real root of $f(x)$ in interval $(-1, 0)$ because $f(0)$ and $f(-1)$ are of opposite signs. But I would like to present a rather elementary argument for the uniqueness of the root. Clearly we can see that if $x > 0$ then $f(x) > 0$, so that the root if any must be negative. Let's assume on the contrary that there are two distinct roots of $f(x)$ namely $a, b$ and $a \neq b$ and both $a, b$ are negative. Then $f(a) = 0 = f(b)$. So we have $f(a) - f(b) = 0$. This means that $$\begin{aligned}4(a^{5} - b^{5}) + (a^{3} - b^{3}) + 2(a - b) &= 0\\
\Rightarrow 4(a^{4} + a^{3}b + a^{2}b^{2} + ab^{3} + b^{4}) + (a^{2} + ab + b^{2}) + 2 &= 0\end{aligned}$$ Now we can see that both $a, b$ are negative and hence each term of LHS is positive and hence can't sum to $0$ which is on RHS. This contradiction shows that we can't have two distinct roots $a, b$ of $f(x)$.
