Question

question 12 (1 point)
let ( g(x) ) be a function with the following properties:
( g(2)=5, g^{prime}(2)=12, g^{prime prime}(2)=8 ).
if ( f(x)=x^{3}-g(x) ), then which of the following is always true?
( f(2) ) is an absolute minimum
( f(2) ) is a local maximum
( f(2) ) is a singular point
( f(2) ) is a local minimum
( f(2) ) is an absolute maximum

Explanation:

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

Using the power rule \((x^n)^\prime=nx^{n - 1}\) and the difference rule \((u - v)^\prime=u^\prime - v^\prime\).
If \(f(x)=x^{3}-g(x)\), then \(f^\prime(x)=3x^{2}-g^\prime(x)\).
Evaluate \(f^\prime(x)\) at \(x = 2\): \(f^\prime(2)=3\times(2)^{2}-g^\prime(2)\).
Substitute \(g^\prime(2)=12\) into the equation: \(f^\prime(2)=3\times4 - 12=12 - 12=0\).

Step2: Find the second - derivative of \(f(x)\)

Using the power rule \((x^n)^\prime=nx^{n - 1}\) and the difference rule \((u - v)^\prime=u^\prime - v^\prime\) again.
If \(f^\prime(x)=3x^{2}-g^\prime(x)\), then \(f^{\prime\prime}(x)=6x - g^{\prime\prime}(x)\).
Evaluate \(f^{\prime\prime}(x)\) at \(x = 2\): \(f^{\prime\prime}(2)=6\times2 - g^{\prime\prime}(2)\).
Substitute \(g^{\prime\prime}(2)=8\) into the equation: \(f^{\prime\prime}(2)=12 - 8 = 4>0\).

Step3: Use the second - derivative test

The second - derivative test states that if \(f^\prime(c)=0\) and \(f^{\prime\prime}(c)>0\), then \(f(x)\) has a local minimum at \(x = c\).
Since \(f^\prime(2)=0\) and \(f^{\prime\prime}(2)=4>0\), \(f(x)\) has a local minimum at \(x = 2\).

Answer:

\(f(2)\) is a local minimum