Question

determine the open intervals on which the graph of the function is concave upward or concave downward. (enter your answers using interval notation. if an answer does not exist, enter dne.)

( g ( x ) = 9 x ^ { 2 } - x ^ { 3 } )

concave upward

concave downward

Explanation:

Step1: Find the first derivative

Using the power rule \((x^n)^\prime = nx^{n - 1}\), for \(g(x)=9x^{2}-x^{3}\), we have \(g^\prime(x)=\frac{d}{dx}(9x^{2}-x^{3})=18x - 3x^{2}\).

Step2: Find the second derivative

Differentiate \(g^\prime(x)\) again. \(g^{\prime\prime}(x)=\frac{d}{dx}(18x - 3x^{2})=18-6x\).

Step3: Find the inflection point

Set \(g^{\prime\prime}(x) = 0\), so \(18-6x=0\). Solving for \(x\):

$$
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Step4: Test intervals

• For the interval \((-\infty,3)\), let \(x = 0\). Then \(g^{\prime\prime}(0)=18-6\times0 = 18>0\).
• For the interval \((3,\infty)\), let \(x = 4\). Then \(g^{\prime\prime}(4)=18-6\times4=18 - 24=-6<0\).

Answer:

concave upward: \((-\infty,3)\)
concave downward: \((3,\infty)\)