Question

use a table to graph (f(x)=2x^{3}-4x^{2}-3x + 4) on a separate sheet of paper. then estimate the x - coordinates at which relative maxima and relative minima occur. round your answer to the nearest tenth. the relative maximum occurs between the integers (x=) and (x=). the relative minimum occurs between the integers (x=) and (x=). need help with this question? get a hint. watch: maximum and minimum points

Explanation:

Step1: Find the derivative

First, find the derivative of $f(x)=2x^{3}-4x^{2}-3x + 4$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=6x^{2}-8x - 3$.

Step2: Set the derivative equal to zero

Set $f'(x)=0$, so $6x^{2}-8x - 3 = 0$. Use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 6$, $b=-8$, and $c=-3$. Then $x=\frac{8\pm\sqrt{(-8)^{2}-4\times6\times(-3)}}{2\times6}=\frac{8\pm\sqrt{64 + 72}}{12}=\frac{8\pm\sqrt{136}}{12}=\frac{8\pm2\sqrt{34}}{12}=\frac{4\pm\sqrt{34}}{6}$.

Step3: Approximate the values

$\sqrt{34}\approx5.83$. So $x_1=\frac{4+\sqrt{34}}{6}\approx\frac{4 + 5.83}{6}=\frac{9.83}{6}\approx1.6$ and $x_2=\frac{4-\sqrt{34}}{6}\approx\frac{4-5.83}{6}=\frac{-1.83}{6}\approx - 0.3$.

The relative maximum occurs between the integers $x=-1$ and $x = 0$. The relative minimum occurs between the integers $x = 1$ and $x = 2$.

Answer:

The relative maximum occurs between the integers $x=-1$ and $x = 0$.
The relative minimum occurs between the integers $x = 1$ and $x = 2$.