Question

use a table to graph f(x)=x^4 + 2x - 1 on a separate sheet of paper. then estimate the x - coordinates at which relative maxima and relative minima occur. the relative maximum occurs near x = select choice. the relative minimum occurs near x = select choice.

Explanation:

Step1: Find the derivative

The derivative of $f(x)=x^{4}+2x - 1$ is $f'(x)=4x^{3}+2$.

Step2: Set the derivative equal to zero

Set $4x^{3}+2 = 0$. Then $4x^{3}=-2$, so $x^{3}=-\frac{1}{2}$, and $x =-\frac{1}{\sqrt[3]{2}}\approx - 0.79$.

Step3: Analyze the second - derivative

The second - derivative $f''(x)=12x^{2}$. When $x =-\frac{1}{\sqrt[3]{2}}$, $f''(-\frac{1}{\sqrt[3]{2}})=12\times(-\frac{1}{\sqrt[3]{2}})^{2}>0$, so there is a relative minimum at $x =-\frac{1}{\sqrt[3]{2}}$. Since the function is a quartic function with a positive leading coefficient ($a = 1$ in $y=x^{4}+2x - 1$), there is no relative maximum.

Answer:

The relative maximum occurs near $x$: None
The relative minimum occurs near $x\approx - 0.79$