the graph of the following function has one relative extreme point. find it and determine whether it is a…

Explanation:

Step1: Find the first - derivative

Given $f(x)=6x^{2}-5x - 1$, then $f^\prime(x)=\frac{d}{dx}(6x^{2}-5x - 1)=12x - 5$.

Step2: Find the critical points

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

Step3: Find the second - derivative

$f^{\prime\prime}(x)=\frac{d}{dx}(12x - 5)=12$.

Step4: Determine the nature of the critical point

Since $f^{\prime\prime}(\frac{5}{12}) = 12>0$, the function has a relative minimum at $x = \frac{5}{12}$. To find the $y$ - value, substitute $x=\frac{5}{12}$ into $f(x)$: $f(\frac{5}{12})=6(\frac{5}{12})^{2}-5(\frac{5}{12})-1=6\times\frac{25}{144}-\frac{25}{12}-1=\frac{25}{24}-\frac{50}{24}-\frac{24}{24}=-\frac{49}{24}$.

Answer:

The relative extreme point is $(\frac{5}{12},-\frac{49}{24})$, and it is a relative minimum.