Question

the graph of the function has one relative extreme point. plot this point and check the concavity there. using only this information, sketch the graph

$f(x)=4x^{2}-5$

the relative extreme point on the graph is (type an ordered pair.)

Explanation:

Step1: Find the derivative

Differentiate $f(x)=4x^{2}-5$ using the power - rule. The derivative $f'(x)=\frac{d}{dx}(4x^{2}-5)=8x$.

Step2: Find the critical points

Set $f'(x) = 0$. So, $8x=0$, which gives $x = 0$.

Step3: Find the y - coordinate of the critical point

Substitute $x = 0$ into the original function $f(x)$. Then $f(0)=4(0)^{2}-5=-5$.

Step4: Find the second - derivative

Differentiate $f'(x)=8x$ to get the second - derivative $f''(x)=\frac{d}{dx}(8x)=8$.

Step5: Check the concavity

Since $f''(x)=8>0$, the function is concave up at $x = 0$.

Answer:

$(0, - 5)$