16. show that y = x^3 + 4x + 16 has exactly one x intercept.
16. show that y = x^3 + 4x + 16 has exactly one x intercept.
Answer
Explanation:
Step1: Find the derivative
The derivative of $y = x^{3}+4x + 16$ using the power - rule $(x^n)'=nx^{n - 1}$ is $y'=3x^{2}+4$. Since $x^{2}\geq0$ for all real $x$, then $3x^{2}\geq0$ and $y'=3x^{2}+4\geq4>0$ for all real $x$. So the function $y = x^{3}+4x + 16$ is strictly increasing on the entire real line $(-\infty,\infty)$.
Step2: Use the Intermediate Value Theorem
Evaluate the function at $x=-2$: $y(-2)=(-2)^{3}+4\times(-2)+16=-8 - 8 + 16=0$. Since the function is strictly increasing (it is always going up as $x$ increases), it can cross the $x$-axis only once.
Answer:
The function $y = x^{3}+4x + 16$ is strictly increasing (because $y'>0$ for all $x\in R$) and has an $x$-intercept at $x = - 2$. So it has exactly one $x$-intercept.