Question

second derivative test: problem 1
(1 point)
consider the function ( f(x)=5x + 8x^{-1} ).
note that this function has no inflection points, but ( f(x) ) is undefined at ( x = b ) where
( b=)
for each of the following intervals, tell whether ( f(x) ) is concave up (type in cu) or concave down (type in cd).
( (-infty,b):)
( (b,infty):)
note: you can earn partial credit on this problem.

Explanation:

Step1: Find the second - derivative of \(f(x)\)

First, find the first - derivative using the power rule \((x^n)^\prime=nx^{n - 1}\).
For \(f(x)=5x + 8x^{-1}\), \(f^\prime(x)=5-8x^{-2}=5-\frac{8}{x^{2}}\).
Then find the second - derivative: \(f^{\prime\prime}(x)=\frac{16}{x^{3}}\).

Step2: Find the value of \(B\)

The second - derivative \(f^{\prime\prime}(x)=\frac{16}{x^{3}}\) is undefined when \(x = 0\), so \(B = 0\).

Step3: Determine the concavity on \((-\infty,0)\)

Let \(x=-1\) (a test point in the interval \((-\infty,0)\)). Then \(f^{\prime\prime}(-1)=\frac{16}{(-1)^{3}}=-16<0\).
So \(f(x)\) is concave down (CD) on \((-\infty,0)\).

Step4: Determine the concavity on \((0,\infty)\)

Let \(x = 1\) (a test point in the interval \((0,\infty)\)). Then \(f^{\prime\prime}(1)=\frac{16}{(1)^{3}}=16>0\).
So \(f(x)\) is concave up (CU) on \((0,\infty)\).

Answer:

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