Question

consider the function ( f(x)=4x + 9x^{-1} ). note that this function has no inflection points, but ( f(x) ) is undefined at ( x = b ) where ( b=square ) for each of the following intervals, tell whether ( f(x) ) is concave up (type in cu) or concave down (type in cd). ( (-infty,b):square ) ( (b,infty):square )

Explanation:

Step1: Find the first - derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(y = f(x)=4x + 9x^{-1}\), we have \(f^\prime(x)=\frac{d}{dx}(4x)+\frac{d}{dx}(9x^{-1})\).
\(f^\prime(x)=4-9x^{-2}=4-\frac{9}{x^{2}}\).

Step2: Find the second - derivative

Differentiate \(f^\prime(x)\) with respect to \(x\). Using the power rule again, \(f^{\prime\prime}(x)=\frac{d}{dx}(4)-\frac{d}{dx}(9x^{-2})\).
\(f^{\prime\prime}(x)=0 + 18x^{-3}=\frac{18}{x^{3}}\).

Step3: Find where \(f^{\prime\prime}(x)\) is undefined

The function \(y = f^{\prime\prime}(x)=\frac{18}{x^{3}}\) is undefined when \(x = 0\), so \(B = 0\).

Step4: Test the concavity on the intervals

• For the interval \((-\infty,0)\):

Let \(x=-1\), then \(f^{\prime\prime}(-1)=\frac{18}{(-1)^{3}}=-18<0\). So \(f(x)\) is concave down (CD) on \((-\infty,0)\).

• For the interval \((0,\infty)\):

Let \(x = 1\), then \(f^{\prime\prime}(1)=\frac{18}{1^{3}}=18>0\). So \(f(x)\) is concave up (CU) on \((0,\infty)\).

Answer:

\(B = 0\); \((-\infty,B)\): CD; \((B,\infty)\): CU