Question

given ( f(x)=x^{4}-32x + 4 ), find the following.
a) critical point(s): ( x=)
b) interval(s) of increasing:
interval(s) of decreasing:
c) ( f(x) ) has a select an answer at ( x=)

Explanation:

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

The derivative of \(f(x)=x^{4}-32x + 4\) using the power rule \((x^{n})^\prime=nx^{n - 1}\) is \(f^\prime(x)=4x^{3}-32\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(4x^{3}-32=0\).
Factor out \(4\): \(4(x^{3}-8)=0\), then \(x^{3}-8 = 0\).
Using the formula \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\), with \(a=x\) and \(b = 2\), we have \((x - 2)(x^{2}+2x + 4)=0\).
The quadratic \(x^{2}+2x + 4\) has discriminant \(\Delta=(2)^{2}-4\times4=-12<0\). So the only real - valued solution is \(x = 2\) from \(x-2=0\).

Step3: Determine the intervals of increase and decrease

We use test points.
Let's choose a test point in the interval \((-\infty,2)\), say \(x = 0\). Then \(f^\prime(0)=4\times0^{3}-32=-32<0\).
Let's choose a test point in the interval \((2,\infty)\), say \(x = 3\). Then \(f^\prime(3)=4\times3^{3}-32=4\times27-32=108 - 32 = 76>0\).
So \(f(x)\) is decreasing on the interval \((-\infty,2)\) and increasing on the interval \((2,\infty)\).

Step4: Determine the nature of the critical point

Since \(f(x)\) changes from decreasing \((f^\prime(x)<0)\) to increasing \((f^\prime(x)>0)\) at \(x = 2\), by the first - derivative test, \(f(x)\) has a local minimum at \(x = 2\).

Answer:

a) \(x = 2\)
b) Interval(s) of increasing: \((2,\infty)\)
Interval(s) of decreasing: \((-\infty,2)\)
c) \(f(x)\) has a local minimum at \(x = 2\)