Question

the graph pictured to the right is the graph of ( f(x) ), the derivative of a polynomial function, ( f(x) ). which of the following statements is/are true?
i. ( f(x) ) is increasing on the interval ( (-5,-2) ) and ( (-2,2) ).
ii. ( f(x) ) has a relative minimum when ( x=-5 ).
iii. the slope of the normal line drawn to ( f(x) ) at ( x=-4 ) is undefined.

Explanation:

Step1: Analyze the sign of \(f^{\prime}(x)\) for increasing - decreasing

A function \(y = f(x)\) is increasing when \(f^{\prime}(x)>0\).
From the graph of \(y = f^{\prime}(x)\), in the interval \((-5,-2)\), \(f^{\prime}(x)>0\). But at \(x=-2\), \(f^{\prime}(x) = 0\). In the interval \((-2,2)\), \(f^{\prime}(x)>0\). So \(f(x)\) is increasing on \((-5,2)\) (excluding any points where \(f^{\prime}(x)\) is not well - defined in the open interval). So statement I is True.

Step2: Check for relative minimum at \(x = - 5\)

A relative minimum of \(y = f(x)\) occurs when \(f^{\prime}(x)\) changes sign from negative to positive.
For \(x=-5\), we consider the behavior of \(f^{\prime}(x)\) around \(x = - 5\). If \(x<-5\), \(f^{\prime}(x)<0\) (assuming the domain to the left of \(x=-5\) is such that we can analyze the sign change). As \(x\) moves from the left of \(x=-5\) to the right of \(x=-5\), \(f^{\prime}(x)\) changes from negative to positive. So \(f(x)\) has a relative minimum at \(x=-5\) by the First - Derivative Test. So statement II is True.

Step3: Find the slope of the normal line at \(x=-4\)

The slope of the tangent line to \(y = f(x)\) at \(x = a\) is \(m_{t}=f^{\prime}(a)\). The slope of the normal line \(m_{n}\) is given by \(m_{n}=-\frac{1}{m_{t}}\) (when \(m_{t}
eq0\)).
At \(x=-4\), \(f^{\prime}(-4)=3\) (from the point \((-4,3)\) on the graph of \(y = f^{\prime}(x)\)). The slope of the tangent line \(m_{t}=3\), and the slope of the normal line \(m_{n}=-\frac{1}{3}\). So statement III is False.

Answer:

I. True, II. True, III. False