Answer
a. (i) $y=0$, $x=\pm2$; $y'=0$, $x=0$.
(ii) $y=0$, $x=-5,-3$; $y'=0$, $x=-4$.
(iii) $y=0$, $x=-1,2$; $y'=0$, $x=0,2$.
(iv) $y=0$, $x=0,9,24$; $y'=0$, $x=4,18$. See figure.
b. See explanations.
Work Step by Step
a. (i) Given $y=x^2-4$, we have $y'=2x$. Letting $y=0$, we get $x=\pm2$. Letting $y'=0$, we get $x=0$. We can plot $x=\pm2, 0$ on a line as shown.
(ii) Given $y=x^2+8x+15=(x+5)(x+3)$, we have $y'=2x+8$. Letting $y=0$, we get $x=-5,-3$. Letting $y'=0$, we get $x=-4$. We can plot $x=-5,-4,-3$ on a line as shown.
(iii) Given $y=x^3-3x^2+4=(x+1)(x-2)^2$, we have $y'=3x^2-6x$. Letting $y=0$, we get $x=-1,2$. Letting $y'=0$, we get $x=0,2$. We can plot $x=-1,0,2$ on a line as shown.
(iv) Given $y=x^3-33x^2+216x=x(x-9)(x-24)$, we have $y'=3x^2-66x+216=3(x^2-22x+72)=3(x-4)(x-18)$. Letting $y=0$, we get $x=0,9,24$. Letting $y'=0$, we get $x=4,18$. We can plot $x=0,4,9,18,24$ on a line as shown.
b. Rolle's Theorem states that if $f(a)=f(b)$ then there is at least one number $c$ in $(a,b)$ at which $f'(c)=0$ (here function $f$ is continuous over $[a,b]$ and differentiable in $(a,b)$). Given a polynomial function $f(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$, we have $f'(x)=nx^{n-1}+(n-1)a_{n-1}x^{n-2}+...+a_1$. Assume there are two zeros $x=a,b$ in $f$; we have $f(a)=f(b)=0$ and based on the Rolle's Theorem, there is a number $c$ between $a,b$ such that $f'(c)=0$. Here the polynomial function is continuous and differentiable on the entire domain.
