given the function f(x)=2x³/3 - x² + x/2 + 1/4, answer the following. f(x) has a relative maximum of…

Explanation:

Step1: Find the first - derivative

The function is $f(x)=\frac{2x^{3}}{3}-x^{2}+\frac{x}{2}+\frac{1}{4}$. Using the power rule $\frac{d}{dx}(x^{n}) = nx^{n - 1}$, we have $f'(x)=2x^{2}-2x+\frac{1}{2}$.

Step2: Set the first - derivative equal to zero

Set $f'(x)=0$, so $2x^{2}-2x+\frac{1}{2}=0$. Multiply through by 2 to get $4x^{2}-4x + 1=0$. This is a perfect - square trinomial: $(2x - 1)^{2}=0$, and $x=\frac{1}{2}$.

Step3: Find the second - derivative

Differentiate $f'(x)=2x^{2}-2x+\frac{1}{2}$ with respect to $x$. Using the power rule, $f''(x)=4x-2$.

Step4: Evaluate the second - derivative at the critical point

Substitute $x = \frac{1}{2}$ into $f''(x)$. $f''(\frac{1}{2})=4\times\frac{1}{2}-2=0$. Since the second - derivative test is inconclusive, we can rewrite $f(x)$ and analyze the behavior. We can also use the fact that $f(x)=\frac{2x^{3}}{3}-x^{2}+\frac{x}{2}+\frac{1}{4}$, and $f(\frac{1}{2})=\frac{2}{3}\times(\frac{1}{2})^{3}-(\frac{1}{2})^{2}+\frac{1}{2}\times\frac{1}{2}+\frac{1}{4}=\frac{2}{3}\times\frac{1}{8}-\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{1}{12}+\frac{1}{4}=\frac{1 + 3}{12}=\frac{1}{3}$ (There is a mistake above, let's start from step 1 again correctly)

Step1: Find the first - derivative

Given $f(x)=\frac{2x^{3}}{3}-x^{2}+\frac{x}{2}+\frac{1}{4}$, by the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $f'(x)=2x^{2}-2x+\frac{1}{2}$.

Step2: Set $f'(x) = 0$

$2x^{2}-2x+\frac{1}{2}=0$. Multiply through by 2: $4x^{2}-4x + 1 = 0$. Factoring gives $(2x - 1)^{2}=0$, so $x=\frac{1}{2}$.

Step3: Find the second - derivative

Differentiate $f'(x)$: $f''(x)=4x - 2$.

Step4: Evaluate the second - derivative at the critical point

Substitute $x=\frac{1}{2}$ into $f''(x)$: $f''(\frac{1}{2})=4\times\frac{1}{2}-2=0$. We can also use the first - derivative test. Let's choose test points. For $x<\frac{1}{2}$, say $x = 0$, then $f'(0)=\frac{1}{2}>0$. For $x>\frac{1}{2}$, say $x = 1$, then $f'(1)=2 - 2+\frac{1}{2}=\frac{1}{2}>0$. Since the sign of $f'(x)$ does not change around $x=\frac{1}{2}$, the function has no relative maximum.

Answer:

E. f(x) has no relative maximum