Question

Determine the interval in which \(f(x)\) is increasing if

\({f^\prime }(x) = {(x + 1)^2}{(x - 3)^5}{(x - 6)^4}\).

Short answer

The resultant answer is for \(a < 0\), the graph will be shifted \(4{a^3}\) units down.

Step-by-step solution

Given data

The given function is \({f^\prime }(x) = {(x + 1)^2}{(x - 3)^5}{(x - 6)^4}\).

Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

Simplify the expression

The expression is;

\(\begin{array}{c}{y^\prime } = 3{x^2} - 3{a^2},{\rm{ so}}\\{y^\prime } = 0{\rm{ when }}x = \pm a\end{array}\)

Therefore;

\(y\)is increasing when \(x \in ( - \infty , - |a|) \cup (|a|,\infty )\) and decreasing when \(x \in ( - |a|,|a|)\)

\(\begin{array}{c}{y^{\prime \prime }} = 6x\\{y^{\prime \prime }} = 0{\rm{ when }}x = 0\end{array}\)

Therefore;

\(f\) is concave upward when \(x > 0\) and concave downward when \(x < 0\) Moreover \(y = 0\) when \(x = a\) or \(x = - 2a\).

The expression is;

\(\begin{array}{c}{y^\prime } = 3{x^2} - 3{a^2},{\rm{ so}}\\{y^\prime } = 0{\rm{ when }}x = \pm a\end{array}\)

Therefore;

\(y\)is increasing when \(x \in ( - \infty , - |a|) \cup (|a|,\infty )\) and decreasing when \(x \in ( - |a|,|a|)\)

\(\begin{array}{c}{y^{\prime \prime }} = 6x\\{y^{\prime \prime }} = 0{\rm{ when }}x = 0\end{array}\)

Therefore;

\(f\) is concave upward when \(x > 0\) and concave downward when \(x < 0\) Moreover \(y = 0\) when \(x = a\) or \(x = - 2a\).

The graph above is for \(a > 0\).

For \(a = 0\), we have a special case since the function will be just \(y = {x^3}\) and thus will have no local minima or maxima. For \(a < 0\), the graph will be shifted \(4{a^3}\) units down.