16. show that $y = x^{3}+4x + 16$ has exactly one $x$ intercept.

Explanation:

Step1: Find the derivative

The derivative of $y = x^{3}+4x + 16$ using the power - rule $(x^n)^\prime=nx^{n - 1}$ is $y^\prime=3x^{2}+4$. Since $x^{2}\geq0$ for all real $x$, then $3x^{2}\geq0$ and $y^\prime=3x^{2}+4\geq4>0$ for all real $x$. So the function $y = x^{3}+4x + 16$ is a strictly increasing function.

Step2: Analyze the limits

$\lim_{x\rightarrow-\infty}(x^{3}+4x + 16)=-\infty$ and $\lim_{x\rightarrow+\infty}(x^{3}+4x + 16)=+\infty$. Because the function $y = f(x)=x^{3}+4x + 16$ is continuous (as a polynomial function) and strictly increasing, and it goes from negative to positive values as $x$ varies from $-\infty$ to $+\infty$, by the Intermediate Value Theorem, there must be exactly one value of $x$ for which $y = 0$.

Answer:

The function $y=x^{3}+4x + 16$ is a continuous, strictly - increasing function (since $y^\prime>0$ for all $x$) that goes from $-\infty$ to $+\infty$ as $x$ goes from $-\infty$ to $+\infty$. By the Intermediate Value Theorem, it has exactly one $x$ - intercept.