Question

using technology to help analyze a function
graph the function y = x³ + x² - 6x + 5. what are the coordinates, to the nearest tenth, of the relative maximum of the graph?
(1.5, 1.3)
(-1.8, 13.2)
(-2.2, 15.8)
(1.1, 0.9)

Explanation:

Step1: Differentiate the function

Differentiate $y = x^{3}+x^{2}-6x + 5$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we get $y'=3x^{2}+2x - 6$.

Step2: Find the critical points

Set $y' = 0$, so $3x^{2}+2x - 6=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$ (here $a = 3$, $b = 2$, $c=-6$), we have $x=\frac{-2\pm\sqrt{2^{2}-4\times3\times(-6)}}{2\times3}=\frac{-2\pm\sqrt{4 + 72}}{6}=\frac{-2\pm\sqrt{76}}{6}=\frac{-2\pm2\sqrt{19}}{6}=\frac{-1\pm\sqrt{19}}{3}$.

Step3: Determine the nature of the critical points

Differentiate $y'$ to get the second - derivative $y'' = 6x+2$. Substitute the critical points into $y''$. For $x=\frac{-1+\sqrt{19}}{3}\approx1.1$, $y''>0$, so it is a relative minimum. For $x=\frac{-1 - \sqrt{19}}{3}\approx - 1.8$, $y''<0$, so it is a relative maximum.
Substitute $x\approx - 1.8$ into the original function $y=(-1.8)^{3}+(-1.8)^{2}-6\times(-1.8)+5=-5.832 + 3.24+10.8 + 5=13.208\approx13.2$.

Answer:

B. $(-1.8,13.2)$