### Applications of differentiation: Analysis of functions

### Monotonicity

Monotony

Let #f# be a function which is defined on an interval #I#.

- #f# is called
**increasing**if, for all numbers of #c#, #d# in #I# with #c\lt d#, we have #f(c)\lt f(d)#. - #f# is called
**decreasing**if, for all numbers #c#, #d# in #I# with #c\lt d#, we have #f(c)\gt f(d)#. - #f# is called
**weakly increasing**if, for all numbers #c#, #d# in #I# with #c\lt d#, we have #f(c)\le f(d)#. - #f# is called
**weakly decreasing**if, for all numbers #c#, #d# in #I# with #c\lt d#, we have #f(c)\ge f(d)#.

A function that is only increasing or only decreasing is called **monotonic**.

The graph of an increasing function moves to the top right.

By looking at the derivative of a function we can find out if it is increasing or decreasing.

Derivative criterion for monotony

Assume that #f# is a continuous function on #I# which is differentiable on the open interval #\ivoo{a}{b}# where #a# and #b# are the boundary points of #I#.

- If #f'(x)\gt0# for all #x\in\ivoo{a}{b}#, then #f# is increasing on #I#.
- If #f'(x)\lt0# for all #x\in\ivoo{a}{b})#, then #f# is decreasing on #I#.
- If #f'(x)\ge0# for all #x\in\ivoo{a}{b}#, then #f# is weakly increasing on #I#.
- If #f'(x)\le0# for all #x\in\ivoo{a}{b}#, then #f# is weakly decreasing on #I#.

Consider the function \(f(x) = x^4\cdot \euler^ {- 4\cdot x } \). On which interval(s) is this function increasing?

Write your answer in the form #a \leq x \leq b# if there is only one interval, or #(a \leq x \leq b) \lor ( c \leq x \leq d)# in the case of two intervals. Here, #a#, #b#, #c#, and #d# are exact numbers.

Use, if necessary, the values of the derivative from the sign diagram below.

Write your answer in the form #a \leq x \leq b# if there is only one interval, or #(a \leq x \leq b) \lor ( c \leq x \leq d)# in the case of two intervals. Here, #a#, #b#, #c#, and #d# are exact numbers.

Use, if necessary, the values of the derivative from the sign diagram below.

#x# | #-0.5# | #0.5# | #1.0# | #2.0# |

#f'(x)# | #-5.54# | #0.03# | #0.00# | #-0.01# |

Solution #0\leq x \leq 1#

The function #f# is increasing on an interval #I# if #f'(x) \ge 0# for all #x# in #I#. The first derivative of the function is: \[ f'(x)= 4\cdot x^3\cdot \euler^ {- 4\cdot x }-4\cdot x^4\cdot \euler^ {- 4\cdot x } \tiny. \] By moving #x-1# and #x^3# outside of the brackets, we can rewrite this to \[f'(x)=-4\cdot \left(x-1\right)\cdot x^3\cdot \euler^ {- 4\cdot x }\tiny.\] The graph of this derivative intersects the #x#-axis in #x=0# and #x=1#. Now we take a look at the sign diagram to determine the sign of the derivative on both sides of the points #x=0# and #x=1#. For #x \ge 0# and #x \le 1#, the value #f'(x)# is #\text{positive}#, so #f(x)# is increasing on #\ivcc{0}{1}#; in other words, for # 0\leq x \leq 1#.

Below the graph of the function #x^4\cdot \euler^ {- 4\cdot x }# is drawn.

The function #f# is increasing on an interval #I# if #f'(x) \ge 0# for all #x# in #I#. The first derivative of the function is: \[ f'(x)= 4\cdot x^3\cdot \euler^ {- 4\cdot x }-4\cdot x^4\cdot \euler^ {- 4\cdot x } \tiny. \] By moving #x-1# and #x^3# outside of the brackets, we can rewrite this to \[f'(x)=-4\cdot \left(x-1\right)\cdot x^3\cdot \euler^ {- 4\cdot x }\tiny.\] The graph of this derivative intersects the #x#-axis in #x=0# and #x=1#. Now we take a look at the sign diagram to determine the sign of the derivative on both sides of the points #x=0# and #x=1#. For #x \ge 0# and #x \le 1#, the value #f'(x)# is #\text{positive}#, so #f(x)# is increasing on #\ivcc{0}{1}#; in other words, for # 0\leq x \leq 1#.

Below the graph of the function #x^4\cdot \euler^ {- 4\cdot x }# is drawn.

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