Q42E - Campbell Biology

Question

The graph of a function \(y = f\left( x \right)\) is shown. At which point(s) are the following true?

 

(a) \(\frac{{dy}}{{dx}}\) and \(\frac{{{d^2}y}}{{d{x^2}}}\) are both positive.

(b) \(\frac{{dy}}{{dx}}\) and \(\frac{{{d^2}y}}{{d{x^2}}}\) are both negative.

(c) \(\frac{{dy}}{{dx}}\) is negative but \(\frac{{{d^2}y}}{{d{x^2}}}\) is positive.

Step-by-Step Solution

Answer

a)

 

The condition \(\frac{{dy}}{{dx}} > 0\) and \(\frac{{{d^2}y}}{{d{x^2}}} > 0\) is true at point B on the graph.

 

b) 

 

The condition \(\frac{{dy}}{{dx}} < 0\) and \(\frac{{{d^2}y}}{{d{x^2}}} < 0\) is true at point E on the graph.

 

c)

 

The condition \(\frac{{dy}}{{dx}} < 0\) and \(\frac{{{d^2}y}}{{d{x^2}}} > 0\) is true at point A . 

It is observed that \(\frac{{dy}}{{dx}} > 0\) and \(\frac{{{d^2}y}}{{d{x^2}}} < 0\) at the point \(C\), \(\frac{{dy}}{{dx}} = 0\) and \(\frac{{{d^2}y}}{{d{x^2}}} \le 0\) at the point \(D\).

Step 1: Increasing/ Decreasing Test and Concavity Test

The increasing and decreasing test as shown below:

 

1. The function \(f\) is increasing on the interval when \(f'\left( x \right) > 0\) on an interval.
2. The function \(f\) is decreasing on the interval when \(f'\left( x \right) < 0\) on an interval.

 

The concavity test as shown below:

 

1. When \(f''\left( x \right) > 0\) on an interval \(I\) then the graph of \(f\) is said to be concave upward on \(I\).
2. When \(f''\left( x \right) < 0\) on an interval \(I\) then the graph of \(f\) is said to be concave downward on \(I\).

Step 2: Determine at what points the condition are true

The explanation for the given condition is shown below:

 

(a)

 

\(\frac{{dy}}{{dx}}\) and \(\frac{{{d^2}y}}{{d{x^2}}}\) are both positive. This follows that \(\frac{{dy}}{{dx}} > 0\) (show that \(f\) is increasing) and \(\) (shows that \(f\) is concave upward) is true at the point \(B\).

 

b)

 

\(\frac{{dy}}{{dx}}\) and \(\frac{{{d^2}y}}{{d{x^2}}}\) are both negative. This follows that \(\frac{{dy}}{{dx}} < 0\) (show that 

\(f\) is decreasing) and \(\frac{{{d^2}y}}{{d{x^2}}} < 0\) (shows that \(f\) is concave downward) is true at the point \(E\).

 

c)

 

\(\frac{{dy}}{{dx}}\) is negative but \(\frac{{{d^2}y}}{{d{x^2}}}\) is positive. This follows that \(\frac{{dy}}{{dx}} < 0\) (the function \(f\) is decreasing) and \(\frac{{{d^2}y}}{{d{x^2}}} > 0\) (the function \(f\) is concave upward) is true at the point \(A\).

 

It is observed that \(\frac{{dy}}{{dx}} > 0\) and \(\frac{{{d^2}y}}{{d{x^2}}} < 0\) at the point \(C\), \(\frac{{dy}}{{dx}} = 0\) and \(\frac{{{d^2}y}}{{d{x^2}}} \le 0\) at the point \(D\).