Question

Suppose the derivative of a function f is:

\(f'\left( x \right) = {\left( {x - {\bf{4}}} \right)^{\bf{2}}}{\left( {x + {\bf{3}}} \right)^{\bf{7}}}{\left( {x - {\bf{5}}} \right)^{\bf{8}}}\)

On what interval(s) is f increasing?

Short answer

f is increasing everywhere in \(\left( { - 3,\infty } \right)\), except \(x = 4\) and \(x = 5\).

Step-by-step solution

Analyze the derivative of f

For the derivative of function \(f'\left( x \right) = {\left( {x - 4} \right)^2}{\left( {x + 3} \right)^7}{\left( {x - 5} \right)^8}\), the terms \({\left( {x - 4} \right)^2}\) and \({\left( {x - 4} \right)^8}\) are always positive. Therefore, the sign of derivative of \(f\left( x \right)\) is decided by the term \({\left( {x + 3} \right)^7}\).

Find the intervals of increasing and decreasing values for f

The sign of the term \({\left( {x + 3} \right)^7}\)is positive for \(x > - 3\). Thus, f is increasing in the intervals \(\left( { - 3,4} \right)\), \(\left( {4,5} \right)\), and \(\left( {5,\infty } \right)\).

The function \(f\left( x \right)\) is continuous in the interval \(\left( { - 3,\infty } \right)\), except \(x = 4\) and \(x = 5\).

Therefore, f is increasing everywhere in \(\left( { - 3,\infty } \right)\), except \(x = 4\) and \(x = 5\).