Question

find the absolute maximum and minimum, if either exists, for ( f(x)=x^{2}-12x + 5 ).
find the first derivative of ( f ).
( f^{prime}(x)=square )
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the absolute minimum is ( square ) at ( x=square ).
b. there is no absolute minimum.
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the absolute maximum is ( square ) at ( x=square ).
b. there is no absolute maximum.

Explanation:

Step1: Find the first derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(f(x)=x^{2}-12x + 5\), we have \(f^\prime(x)=(x^{2})^\prime-(12x)^\prime+(5)^\prime\).
Since \((x^{2})^\prime = 2x\), \((12x)^\prime=12\) and \((5)^\prime = 0\), then \(f^\prime(x)=2x-12\).

Step2: Find critical points

Set \(f^\prime(x)=0\), so \(2x - 12=0\).
Add \(12\) to both sides: \(2x=12\).
Divide both sides by \(2\): \(x = 6\).

Step3: Find the second derivative

Differentiate \(f^\prime(x)=2x - 12\) with respect to \(x\). Using the power rule, \(f^{\prime\prime}(x)=(2x)^\prime-(12)^\prime\).
Since \((2x)^\prime=2\) and \((12)^\prime = 0\), \(f^{\prime\prime}(x)=2>0\).
Since \(f^{\prime\prime}(x)>0\) at \(x = 6\), the function \(f(x)\) has a local (and since it's a parabola \(y=ax^{2}+bx + c,a = 1>0\) an absolute) minimum at \(x = 6\). There is no absolute maximum for the function \(y=x^{2}-12x + 5\) (as \(x\to\pm\infty,y=x^{2}-12x + 5\to+\infty\)).

Answer:

For \(f^\prime(x)\): \(2x-12\).
For the absolute minimum: A. The absolute minimum is at \(x = 6\).
For the absolute maximum: B. There is no absolute maximum.