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.

Short 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-by-step solution

Increasing/ Decreasing Test and Concavity Test

Theincreasing and decreasing testas shown below:

1. The function \(f\) is increasingon 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.

Theconcavity 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 upwardon \(I\).
2. When \(f''\left( x \right) < 0\) on an interval \(I\)then the graph of \(f\) is said to be concave downwardon \(I\).

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 \(D\).