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 (0, - 5) (type an ordered pair.) since the value of f at the relative extreme point is the graph is at this point.

Explanation:

Step1: Find the first - derivative

Differentiate $f(x)=4x^{2}-5$ using the power rule. The power rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. So, $f^\prime(x)=\frac{d}{dx}(4x^{2}-5)=8x$.

Step2: Find the critical points

Set $f^\prime(x) = 0$. So, $8x=0$, which gives $x = 0$. Substitute $x = 0$ into $f(x)$: $f(0)=4(0)^{2}-5=-5$. The relative - extreme point is $(0,-5)$.

Step3: Find the second - derivative

Differentiate $f^\prime(x)=8x$ with respect to $x$. Using the power rule, $f^{\prime\prime}(x)=\frac{d}{dx}(8x)=8$.

Step4: Determine the concavity

Evaluate $f^{\prime\prime}(x)$ at the critical point $x = 0$. Since $f^{\prime\prime}(0)=8>0$, the graph of the function $y = f(x)$ is concave up at the point $(0,-5)$.

Answer:

Since the value of $f^{\prime\prime}$ at the relative extreme point is $8$, the graph is concave up at this point.