Question

find the intervals on which (f(x)) is increasing, the intervals on which (f(x)) is decreasing, and the local extrema. (f(x)=x^{3}+7x - 8). find the derivative (f(x)). (f(x)=3x^{2}+7). enter the function (f(x)) and the derivative (f(x)) into a graphing - calculator to produce a graph of both in the same viewing window. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the function (f(x)) is increasing on . (type your answer using interval notation. type integers or simplified fractions. use a comma to separate answers as needed.) b. the function (f(x)) is never increasing.

Explanation:

Step1: Analyze the sign of the derivative

The derivative is \(f'(x)=3x^{2}+7\). Since \(x^{2}\geq0\) for all real - numbers \(x\), then \(3x^{2}\geq0\) and \(3x^{2}+7\geq7>0\) for all real numbers \(x\).

Step2: Determine increasing and decreasing intervals

A function \(y = f(x)\) is increasing when \(f'(x)>0\) and decreasing when \(f'(x)<0\). Since \(f'(x)=3x^{2}+7>0\) for all \(x\in(-\infty,\infty)\), the function \(f(x)\) is increasing on \((-\infty,\infty)\) and never decreasing.

Step3: Find local extrema

Local extrema occur at points where \(f'(x) = 0\) or \(f'(x)\) is undefined. Since \(f'(x)=3x^{2}+7\) is never equal to \(0\) (because \(3x^{2}+7>0\) for all \(x\)) and is a polynomial (so defined everywhere), there are no local extrema.

Answer:

The function \(f(x)\) is increasing on \((-\infty,\infty)\). The function \(f(x)\) is never decreasing. There are no local extrema.