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 = no relative maximum; the relative minimum occurs near x = 0

Explanation:

Step1: Find the derivative

Differentiate $f(x)=x^{4}+2x - 1$ using the power - rule. The derivative $f^\prime(x)=4x^{3}+2$.

Step2: Set the derivative equal to zero

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

Step3: Analyze the second - derivative

Differentiate $f^\prime(x)$ to get the second - derivative $f^{\prime\prime}(x)=12x^{2}$. Evaluate $f^{\prime\prime}(-\sqrt[3]{\frac{1}{2}})=12(-\sqrt[3]{\frac{1}{2}})^{2}>0$. So the function has a relative minimum at $x =-\sqrt[3]{\frac{1}{2}}$. As $x\to\pm\infty$, $y = f(x)\to+\infty$, so there is no relative maximum.

Answer:

The relative maximum occurs near $x$ = no relative maximum; the relative minimum occurs near $x\approx - 0.79$