the graph of the function has one relative extreme point. plot this point and check the concavity there…

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\n\n$f(x)=4x^{2}-5$\n\nthe relative extreme point on the graph is (type an ordered pair.)
Answer
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:
$(0, - 5)$