# Prove equation has exactly 1 real root

I need to prove that this equation has exactly one real root.

$f(x) = x^3 + 3x^2 + 16$

I have tried proving it by showing that has at least one real root, and then taking the derivative. This method does not work since it can have a negative derivative.

Any hints you could give me would be great.

You have

$f(x) = x^2(x+3) + 16$ so only for $x< -3$ can there be a real root.

Now you have

$f'(x) = 3 x (x+2)$ and for $x< -3$ this is always positive. Hence there is only one real root. This can easily be guessed to be $x = -4$.

$$f'(x)=3x (x+2)$$

$f$ is strictly increasing at $[0,+\infty)$

for $x\ge 0, \; f (x)\ge f (0)=16$.

$f$ is decreasing at $[-2,0]$ thus

for $x\in [-2,0] \; f (x)\ge f (0)$

$f$ is strictly increasing at $(-\infty,-2]$

$f (-2)=20>0$ and $f (-5)=-34 <0$

You conclude that there only one root in $[-5,-2]$.

By Descartes' rule of signs $\,x^3 + 3x^2 + 16\,$ has no positive roots (since there are no sign changes in the sequence of coefficients $\,1, 3, 16\,$), and $1$ negative root (one sign change in $\,-1, 3, 16\,$).

A cubic has three real roots if and only if it has a local max and local min of opposite sign, and only one real root if they have the same sign. For $f(x)=x^3+3x^2+16$, the extrema occur at the zeros of $3x^2+6x$, i.e., at $x=0$ and $x=-2$. Since $f(0)=16$ and $f(-2)=-8+12+16=20$ have the same sign, this cubic has just one real root.

$f'(x) = 3x^2 + 6x = 3x(x+2)$

$f'(x) = 0$ when $x = 0$ or $x=-2$ so those are critical point.

For $x < -2$ then $f'(x) > 0$ and $f(x)$ is increasing. $f(-2) = -8 +12 + 16=20 > 0$. As $x\to -\infty f(x) = x^3 + 3x^2 + 16 \to -\infty$ there is some $k < -2$ so that $f(k) < 0$ and by intermediate value theorem there is a point $c; k < c < -2$ so that $f(c) = 0$. And as $f$ is monotonically increasing on $(-\infty, -2)$ there is only one.

If $-2 < x < 0$ then $f'(x) < 0$ so $f(x)$ is decreasing on $(-2,0)$. $f(0)=16 > 0$ so although $f(x)$ decreases it never decreases below $f(0)=16 > 0$.

And if $x > 0$ then $f'(x) > 0$ and $f(x)$ is increasing. And as it "starts" at $f(0) > 0$ for all $f(x); x >0; f(x) > f(0) > 0$.

• A minor error: $f'(x) = 3x^2 + 6x = 3x(x \color{red}{+} 2)$, which affects the intervals in the remainder of your explanation. – N. F. Taussig Apr 28 '18 at 10:14
• That a serious whoops. – fleablood Apr 28 '18 at 14:39
• Thanks for the diplomacy of calling that a "minor error". It was a minor and careless error to make but the results of the wrong intervals where rather severe. – fleablood Apr 28 '18 at 14:47