Extreme Values One of the important application of the derivative of a function is to find the Extreme Values of a function $f$ in its domain. There are two kind of extreme values, namely Local Extreme Value and Absolute Extreme Value. To understand the complete process, this section is divided in several subsections as follows:
- Critical Points : Problems Q1 ; Q2 ; Q3 ;
- Interval of Increasing and Decreasing Problems Q1 ; Q2 ;
- Local (Absolute) Maximum and Minimum and First Derivative Test Problems Q1 ;
- Concavity of function Problems Q1 ;
- Second Derivative Test for Local Maximum and Minimum Problems Q1 ;
Critical Points
Definition: A critical point $x_0$ of a function $f$ is a point in the domain of the function at which either $f'(x_0)=0$ or $f'(x_0)$ is not defined.
To find Extreme Values , the critical point are very important.To find the Critical Point of a given function following steps are essential:
- Identify the domain of the given function
- Calculate the derivative of $f$ by using appropriate derivative rules/formulas
- Identify (if any) points at which $f'(x)$ is not defined, and validate the points are inside the domain. If yes, then those points are critical points. If the points are not inside the domain, ignore them.
- Identify (if any) points at which $f'(x)=0$, and validate the points are inside the domain. If yes, then those points are critical points. If the points are not inside the domain, ignore them.
Problems related to Critical Points
Q. Find all critical points for $f(x)=x^3+ 6 x^2 +9 x – 8 $
Ans. To find the Critical points, we follow below steps:
- Since $f(x)$ is a polynomial, it is domain in $(-\infty, \infty)$.
- $f'(x) = 3 x^2 + 12 x + 9$
- Since, $f'(x)$ is also a polynomial, it is defined for all $x \in (-\infty, \infty)$.
- Now, $f'(x) =0$ if \[\begin{aligned} 3 x^2 + 12 x + 9 =0 \implies & 3 \bigg(x^2 + 4 x +3 \bigg)=0 \\ \implies& 3 (x+3)(x+1) = 0 \\ \implies& x= -3 \quad \text{and} \quad x=-1. \end{aligned} \]
- Finally, the critical points are $x= -3, – 1$.
Q. Find all critical points for $𝒇(𝒙)=𝟑𝒙^𝟓−𝟏𝟎𝒙^𝟑+𝟏𝟓𝒙 $
Ans. To find the Critical points, we follow below steps:
- Since $f(x)$ is a polynomial, it is domain in $(-\infty, \infty)$.
- $f'(x) = 15 x^4 – 30 x^2 + 15$
- Since, $f'(x)$ is also a polynomial, it is defined for all $x \in (-\infty, \infty)$.
- Now, $f'(x) =0$ if \[\begin{aligned} 15 x^4 – 30 x^2 + 15 =0 \implies & 15 \bigg(x^4 – 2 x^2 + 1 \bigg)=0 \\ \implies& 15 (x^2-1)^2 = 0 \\ \implies& x^2=1 \\ \implies& x=1 \quad \text{and} \quad x=-1. \end{aligned} \]
- Finally, the critical points are $x= 1, – 1$.
Q. Find all critical points for $𝒇(𝒕)=\frac{𝒕+𝟓}{𝒕^𝟐+𝟔𝒕+𝟕} $
Ans. To find the Critical points, we follow below steps:
- Since $f(t)$ is a rational function, it is not defined at $t$ for which $𝒕^𝟐+𝟔𝒕+𝟕 =0 $. (Remark: No need to solve for $t$ here)
- Find the derivative $f'(t)$ using Quotient rule. \[ \begin{aligned} f'(t)& = \frac{(𝒕+𝟓)'(𝒕^𝟐+𝟔𝒕+𝟕)-(𝒕+𝟓)(𝒕^𝟐+𝟔𝒕+𝟕)’ } {(𝒕^𝟐+𝟔𝒕+𝟕)^2}\\ & = \frac{(𝒕^𝟐+𝟔𝒕+𝟕)-(𝒕+𝟓)(2 𝒕 +𝟔) } {(𝒕^𝟐+𝟔𝒕+𝟕)^2}\\ & = \frac{(𝒕^𝟐+𝟔𝒕+𝟕)-(𝒕+𝟓)(2 𝒕^2 + 1𝟔 t + 30) } {(𝒕^𝟐+𝟔𝒕+𝟕)^2}\\ & = \frac{-( 𝒕^2 + 10 t + 23) } {(𝒕^𝟐+𝟔𝒕+𝟕)^2}\\ \end{aligned}\]
- Since, $f'(t)$ is a rational function, it is also not defined for all $t$ such that $𝒕^𝟐+𝟔𝒕+𝟕 =0 $. But as those points are not in the domain, we ignore.
- Now, $f'(t) =0$ if \[\begin{aligned} 𝒕^2 + 10 t + 23 =0 \implies & t= \frac{ -10 \pm \sqrt{10^2 – 4. 1. 23}}{ 2} \\ \implies& t= \frac{-10 \pm \sqrt{8}}{2} \\ \implies& t= \frac{-10 \pm 2\sqrt{2}}{2} \\ \implies& t=- 5 + \sqrt{2} \quad \text{and} \quad t=- 5 – \sqrt{2}. \end{aligned} \]
- Finally, the critical points are $x= – 5 + \sqrt{2}, – 5 – \sqrt{2}$.
Interval of Increasing and Decreasing
- Increasing : A function $f$ is increasing on an interval $\it{I}$ if $f'(x) > 0$ for all $ x \in \it{I}$.
- Decreasing : A function $f$ is decreasing on an interval $\it{I}$ if $f'(x) < 0$ for all $ x \in \it{I}$.
Steps to find the interval of Increasing and decreasing
- Find all Critical points as stated above.
- Divide the domain of the functions into interval at each Critical point. For, example of the domain of the function is $[a, b]$ and $c$, $d$ are two critical points such that $a< c < d < b$. Then, consider the open interval $(a, c)$, $(c, d)$, and $(d, b)$.
- Consider arbitrary point in each interval and find the value of $f’$ at those selected points
- If the sign of $f^{\prime}$ is positive at the selected point, $f$ is increasing on the concern interval, while if $f^{\prime}$ is negative at the selected point, then $f$ is decreasing on the concern interval.
Problems related with finding interval of increasing and decreasing
Q. Find the interval of increasing and decreasing for $f(x)=x^3+ 6 x^2 +9 x – 8 $
Ans. To find the interval of increasing and decreasing, we follow below steps:
- \(f'(x) = 3 x^2 + 12 x + 9 = 3( x^2+ 4 x+ 3) = 3 (x+3)(x+1). \)
- The critical points of $f$ is obtained in earlier section which are $x=-1$ and $x=-3$.
- Now, the remaining steps we will summarized in a tabular form as below
\((-\infty, -3)\) \((-3, -1)\) \((-1, \infty)\) Value of $f^{\prime}$ \(f'(-4) =9\) \(f'(-2) = – 3\) \(f'(2) = 45\) Sign of $f^{\prime}$ \(+ \) \(- \) \(+ \) Conclusion Increasing Decreasing Increasing - The interval of increasing : $(- \infty, -1) \cup (-1, \infty)$
The interval of decreasing : $(-3, -1)$ .
Q. Find the interval of increasing and decreasing for $f(x)= 2 \sqrt{x-2} – \;\dfrac{x}{2} $
Ans. To find the interval of increasing and decreasing, we follow below steps:
- \(f'(x) = \dfrac{1}{\sqrt{x-2}}-\dfrac{1}{2} \)
- The critical points:
- $f'(x)$ is not defined at $\fbox{\textcolor{blue}{x=2}}$.
- The other critical point is obtained by solving $f'(x)=0$. \[ \begin{align*} \implies & \dfrac{1}{\sqrt{x-2}}-\dfrac{1}{2}=0\\ \implies & \sqrt{x-2}= 2 \implies \fbox{\textcolor{blue}{x = 6}} . \end{align*}\]
- Clearly the function $f$ is defined for $[2, \infty)$. Now, the remaining steps we will summarized in a tabular form as below
\((2, 6)\) \((6, \infty)\) Value of $f^{\prime}$ \(f'(3) = \frac{1}{2}\) \(f'(7) = \frac{1}{\sqrt{5}} – \frac{1}{2}\) Sign of $f^{\prime}$ \(+ \) \( – \) Conclusion Increasing Decreasing - The interval of increasing : $(2, 6) $
The interval of decreasing : $(6, \infty)$ .
Local (Absolute) Maximum and Minimum and First Derivative Test
To discuss the extreme values, the graph of a function is very important. A function by its nature of increasing or decreasing, may have lots of ups and down. Accordingly each upper point in certain interval refer as local maximum , while each low point is refer as local minimum of the graph of the function. Thus, through out the domain there may be more than one local maximum and local minimum of a function. This, leads to the concept of the Absolute Maximum and Minimum . Refer to this graph Absolute Max-mini for better picture. Now, it is time for the definition of local and absolute maximum.Definition : Let $f$ be a function defined on a domain $\mathbb{D}$. Let $c$ be a point in the domain $\mathbb{D}$.
- The function $f$ has a Local Maximum at $c$ on an interval $ I \subset \mathbb{D}$ that contains $c$ if $f(x) \leq f(c)$ for all $x \in I$.
- The function $f$ has a Local Minimum at $c$ on an interval $ I \subset \mathbb{D}$ that contains $c$ if $f(x) \geq f(c)$ for all $x \in I$.
- The function $f$ has a Absolute Maximum at $c$ if $f(x) \leq f(c)$ for all $x \in \mathbb{D}$.
- The function $f$ has a Absolute Maximum at $c$ if $f(x) \geq f(c)$ for all $x \in \mathbb{D}$.
Facts to Remember
- If $f$ has a local maximum and minimum at $c$ and \(f^{\prime}\) is defined at $c$, then \(f'(c)=0\).
- If $f$ is continuous on a closed interval, then it has always an absolute maximum and minimum
Finding method of absolute extrema (for continuous functions on closed interval)
- Find all critical points and function value on those points
- Find the value of the functions at the end point
- Compare all values : most highest value is absolute maximum and the smallest one is absolute minimum.
The First Derivative Test for Local Extreme Values
Suppose that $c$ is a critical point of a continuous function $ƒ$, that means $f'(c)=0$ or $f'(c)$ is not defined. Assume that $ƒ$ is differentiable at every point in some interval containing $c$. Then, by moving from left to right across this interval,- if changes from negative to positive at $c$, then ƒ has a local minimum at $c$;
- if changes from negative to positive at $c$, then ƒ has a local minimum at $c$;
- if does not change sign at $c$(both side either positive or negative), then $ƒ$ has no local extremum at $c$.
Problems related with extreme values
Q. Find the local and absolute maximum and minimum values for $f(x)=x^3+ 6 x^2 +9 x – 8 $ in the interval $[- 5, 5]$
Ans. As discussed in earlier sections we have critical points are $x=-1$ and $x=-3$, and from Table we have the interval of increasing and decreasing.
- Since at $x=-3$, there is change from increasing to decreasing, $f$ has a \fbox{local maximum} at $x=-3$; and $ f(-3)= (-3)^3+ 6 (-3)^2 + 9. (-3) -8 = \fbox{-8}$
- Since at $x=-1$, there is change from decreasing to increasing, $f$ has a \fbox{local minimum} at $x=-1$; and $ f(-1)= (-1)^3+ 6 (-1)^2 + 9. (-1) -8 = \fbox{-12}$
- The values at the end points
At $x=-5$, $f(-5) = (-5)^3+ 6 (-5)^2+ 9. (-5) – 8 = – 125 + 150 – 45 -8 = \fbox{ – 28} $
At $x=5$, $f(5) = (5)^3+ 6 (5)^2+ 9. (5) – 8 = 125 + 150 + 45 -8 = \fbox{312} $.
Finally, we have
- Local maximum =$- 8$ at $x=-3$
- Local Minimum = $-12$ at $x=-1$
- Absolute maximum = $312$ at $x=5$
- Absolute minimum= $-28$ at $x=-5$
Concavity of a function
The Graph of a differentiable function $𝑦=𝑓(𝑥)$ is
- Concave up : On an open interval $I$ if $f'(x)$ is increasing on $I$, which is equivalent to say $f^{\prime \prime}(x)> 0$
- Convex Down : On an open interval $I$ if $f'(x)$ is decreasing on $I$, which is equivalent to say $f^{\prime \prime}(x)< 0$.
Problems related with Convexity
Q. Discuss the concavity for $f(x)=x^3+ 6 x^2 +9 x – 8 $.
Ans. First, we need to find derivatives $f’$ and $f”$ .
\[ \begin{align*}
f'(x) &= 3 x^2 + 12 x+ 9 \\
f^{\prime\prime} (x) &= 6x + 12
\end{align*}\]
This implies
$f$ is concave up for $x>-2$ because $f^{\prime\prime}(x) > 0$ when $\fbox{x>-2}$;
$f$ is concave down for $x<-2$ because $f^{\prime\prime}(x) < 0$ when $\fbox{x<-2}$.
The point of inflection is $x=-2$.
The concavity also visible through the graph of $f(x)$.
Second Derivative Test for Local Maximum and Minimum
Suppose that \(f^{\prime\prime}(x)\) is defined on an interval containing a point 𝒄, such that $𝒇'(𝒄)=𝟎$ (i.e. $𝒄$ is a critical point). Then- $𝒇$ has local maximum at $𝒙=𝒄$ if $f^{\prime\prime}(𝒄)<𝟎$ .
- $𝒇$ has local minimum at $𝒙=𝒄$ if $f^{\prime\prime}(𝒄)>𝟎 $.
- Test failed if $f^{\prime\prime}(𝒄)=𝟎$.
Problems related with Local Maximum and Minimum by second derivative test
Q. Discuss Local Maximum and Minimum by second derivative test for $f(x)=x^3+ 6 x^2 +9 x – 8 $.
Ans. First, we need to find derivatives $f’$ and $f”$ .
\[ \begin{align*}
f'(x) &= 3 x^2 + 12 x+ 9 \\
f^{\prime\prime} (x) &= 6x + 12
\end{align*}\]
This implies, the critical points are
\[ f'(x) = 0 \implies 3 (x^2 + 4 x+ 3)=0 \implies x = – 3, -1. \]
Now, by second derivative test, we have
\[ f^{\prime\prime}(-3) = 6. (-3) + 12 = -18 +12 =-6<0 \implies {\bf f \text{\bf has a local maximum at}\; x=-3 }\]
and
\[ f''(-1) = 6. (-1) + 12 = -6 +12 =6>0 \implies {\bf f \text{\bf has a local minimum at}\; x=-1. } \]