Question

For the function \(f\) whose graph is shown, arrange the following numbers in increasing order:

\(0\) \(1\) \({f^'}\left( 2 \right)\) \({f^'}\left( 3 \right)\) \({f^'}\left( 5 \right)\) \({f^{''}}\left( 5 \right)\)

Short answer

The increasing order of the numbers are\({f^{''}}\left( 5 \right) \prec 0 \prec {f^'}\left( 5 \right) \prec {f^'}\left( 2 \right) \prec 1 \prec {f^'}\left( 3 \right)\)

Step-by-step solution

Definition of slope

The slope of a line calculates the "steepness" of a line. It is usually denoted by the letter m. So, the slope of a line is the change in Y divided by the change in X. As the change in Y is very high, the slope can range from zero to any number.

Given Parameters

Given that \(0\) \(1\) \({f^'}\left( 2 \right)\) \({f^'}\left( 3 \right)\) \({f^'}\left( 5 \right)\) \({f^{''}}\left( 5 \right)\)

Explanation for the given function

\(\left( 1 \right)\)\({f^{''}}\left( 5 \right)\) is negative because the graph is concave down at \(x = 5\)

\(\left( 2 \right)\)At \(x = 2,3,5\), the tangents make an acute angle with the positive \(x - \)axis, so the derivatives at all those points are positive.

\(\left( 3 \right)\)At \(x = 3\), the tangent is steeper than \(y = x\)that is for every \(1\) unit increase in \(x\), the change in \(y\) its more than \(1\)unit. Therefore, \({f^'}\left( 3 \right) \succ 1\).

\(\left( 4 \right)\)At \(x = 2,5\), the tangents are less steep than \(y = x\)and the tangent at \(x = 2\)is steeper than it is at \(x = 5\). Therefore, \(0 \prec {f^'}\left( 5 \right) \prec {f^'}\left( 2 \right) \prec 1\)

After combining \(\left( 1 \right),\left( 2 \right),\left( 3 \right),and\left( 4 \right)\)the increasing order of the function is \({f^{''}}\left( 5 \right) \prec 0 \prec {f^'}\left( 5 \right) \prec {f^'}\left( 2 \right) \prec 1 \prec {f^'}\left( 3 \right)\)

Hence the increasing order of the function is \({f^{''}}\left( 5 \right) \prec 0 \prec {f^'}\left( 5 \right) \prec {f^'}\left( 2 \right) \prec 1 \prec {f^'}\left( 3 \right)\)