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)=8 + 2x - 5x^2. the relative extreme point is (1/5, 41/5) (type an ordered pair. simplify your answer. use integers or fractions for any numbers in the expression.) is the relative extreme point a relative maximum or a relative minimum? relative maximum relative minimum

Explanation:

Step1: Find the first - derivative

Given $f(x)=8 + 2x-5x^{2}$, using the power rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=2-10x$.

Step2: Set the first - derivative equal to zero

Set $f'(x) = 0$, so $2-10x=0$. Solving for $x$ gives $10x = 2$, then $x=\frac{1}{5}$.

Step3: Find the $y$ - value of the extreme point

Substitute $x = \frac{1}{5}$ into $f(x)$: $f(\frac{1}{5})=8+2\times\frac{1}{5}-5\times(\frac{1}{5})^{2}=8+\frac{2}{5}-\frac{1}{5}=\frac{40 + 2-1}{5}=\frac{41}{5}$. So the extreme point is $(\frac{1}{5},\frac{41}{5})$.

Step4: Find the second - derivative

Differentiate $f'(x)=2 - 10x$ with respect to $x$. Using the power rule, $f''(x)=-10$.

Step5: Determine if it's a maximum or minimum

Since $f''(\frac{1}{5})=-10<0$, the function has a relative maximum at the point $(\frac{1}{5},\frac{41}{5})$.

Answer:

Relative maximum