Question

the graph of the following function has one relative extreme point. find it and determine whether it is a relative maximum or a relative minimum.
f(x)=6x^{2}-5x - 1
the relative extreme point is
(type an ordered pair. simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find the derivative

The derivative of $f(x)=6x^{2}-5x - 1$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=12x-5$.

Step2: Set the derivative equal to zero

Set $f'(x) = 0$, so $12x-5=0$. Solving for $x$ gives $x=\frac{5}{12}$.

Step3: Find the second - derivative

The second - derivative $f''(x)$ of $f'(x)=12x - 5$ is $f''(x)=12$. Since $f''(\frac{5}{12})=12>0$, the function has a relative minimum at $x = \frac{5}{12}$.

Step4: Find the $y$ - value

Substitute $x=\frac{5}{12}$ into the original function $f(x)=6x^{2}-5x - 1$.
\[

$$\begin{align*} f(\frac{5}{12})&=6\times(\frac{5}{12})^{2}-5\times\frac{5}{12}-1\\ &=6\times\frac{25}{144}-\frac{25}{12}-1\\ &=\frac{25}{24}-\frac{50}{24}-\frac{24}{24}\\ &=\frac{25 - 50-24}{24}\\ &=-\frac{49}{24} \end{align*}$$

\]

Answer:

$(\frac{5}{12},-\frac{49}{24})$