Maths Support: Maxima and minima for differentiable functions on intervals

Question 1

Let $I=[\var{l},\var{m}]$ be an interval and let $g: I \rightarrow I$ be a function defined on this interval
given by :\[g(x) = \simplify{{c}/3*x^3+ {-c*(a+b)}/2*x^2+{c*a*b}*x+{d}}\]

a)

Input the first derivative of $g$ here, factorised into a product of two linear factors in the form $g'(x)=c(x-a)(x-b)$for suitable integers $a$, $b$ and $c$:

$g'(x)=\;\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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b)

Find the stationary points of $g$.

Least stationary point: $\displaystyle{}$ question.score feedback.none Expected answer: Greatest stationary point: $\displaystyle{}$ question.score feedback.none Expected answer:

Do both these stationary points lie in the interval $I$ ?

Gap 2

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c)

Input the second derivative of $g$:

$g''(x)=\;\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

Hence find all local maxima and minima given by the stationary points

Local maximum is at $x=\;\;$ $\displaystyle{}$ question.score feedback.none Expected answer: and the value of the function at the local maximum = $\displaystyle{}$ question.score feedback.none Expected answer:

Local minimum is at $x=\;\;$ $\displaystyle{}$ question.score feedback.none Expected answer: and the value of the function at the local minimum = $\displaystyle{}$ question.score feedback.none Expected answer:

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d)

What are the following values at the end points of the interval $I$ ?

$g(\var{l})=\;\;$ question.score feedback.none Expected answer: $g(\var{m})=\;\;$ question.score feedback.none Expected answer:

Input both to 3 decimal places.

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e)

Global Maximum

At what value of $x \in I$ does $g$ have a global maximum ?

$x=\;\;$ question.score feedback.none Expected answer:

Value of $g$ at this global maximum = question.score feedback.none Expected answer: (input to 3 decimal places).

Global Minimum

At what value of $x \in I$ does $g$ have a global minimum ?

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Advice

Differentiating, we have:

\[g'(x)=\simplify{{c}*x^2+{-c*(a+b)}*x+{c*a*b}={c}*(x+{-a})(x-{b})}\]

Note that we have already factorised the derivative.

Stationary points are given by solving $g'(x)=0 \Rightarrow x=\var{a},\;\;\mbox{or }x=\var{b}$

So the least stationary point is $x=\var{a}$ and the greatest is $x=\var{b}$.

Since $\var{a} > \var{l}$ and $\var{b} \lt \var{m}$ we have that both stationary points are in $I$.

The second derivative is given by \[g''(x)=\simplify{{2*c}*x-{c*(a+b)}}\]

Local Maximum

At the stationary point $x=\var{a}$ we have $g''(\var{a})=\var{c*a-c*b} \lt 0$.

Hence at this value of $x$ we have a local maximum.

The value of the function $g$ at this local maximum is $g(\var{a})= \var{valmax}$.

Local Minimum

At the stationary point $x=\var{b}$ we have $g''(\var{b})=\var{c*b-c*a} \gt 0$.

Hence this point is a local minimum.

The value of the function $g$ at this local minimum is $g(\var{b})= \var{valmin}$.

Global Maximum

First we find the values at the endpoints of the interval $I=[\var{l},\var{m}]$ are:

$g(\var{l})=\var{valbegin}$ to 3 decimal places.

$g(\var{m})=\var{valend}$ to 3 decimal places.

To find the global maximum note that we are only concerned with the values of $g$ on the interval $I$.

So we proceed by comparing the values of the function at the endpoints with the local maximum.

a) If the value at the local maximum is greater than either of the values at the endpoints then this is the global maximum on the interval.

b) Otherwise if the greatest value of the function at the endpoints is greater than the local maximum then this is the global maximum.

\[\begin{array}{c|c|c|c} x & \mbox{Local Maximum}=\var{a} & \var{l} \in I & \var{m} \in I \\ \hline\\ g(x)& \var{valmax} & \var{valbegin} & \var{valend} \\ \end{array} \]

So for our example we see that the global maximum occurs at $x=\var{gma}$ and $g(\var{gma})=\var{valgmax}$.

Global Minimum

We proceed as for the global maximum by comparing the values of the function at the endpoints with the local minimum.

a) If the value at the local minimum is less than either of the values at the endpoints then this is the global minimum on the interval.

b) Otherwise if the least value of the function at the endpoints is less than the local minimum then this is the global minimum.

\[\begin{array}{c|c|c|c} x & \mbox{Local Minimum}=\var{b} & \var{l} \in I & \var{m} \in I \\ \hline\\ g(x)& \var{valmin} & \var{valbegin} & \var{valend} \\ \end{array} \]

In our example we see that the global minimum occurs at $x=\var{gmi}$ and $g(\var{gmi})=\var{valgmin}$.

Question 2

Let $I=[\var{l},\var{m}]$ be an interval and let $g: I \rightarrow \mathbb{R}$ be a function defined on this interval
given by :\[g(x) = \simplify{(x-{a})*(x-{b})^3}\]

Advice

First derivative

Differentiating we have:

\[\begin{eqnarray*} g'(x)&=&\simplify{(x-{b})^3+3*(x-{a})*(x-{b})^2}\\ &=&\simplify{(x-{b})^2(3*(x-{a})+x-{b})}\\ &=&\simplify{4*(x-{k})*(x-{b})^2} \end{eqnarray*} \] and we have factorised the expression.

Stationary points

These are given by solving $g'(x)=0 \Rightarrow x=\var{k},\;\;\mbox{or }x=\var{b}$

Therefore the least stationary point is $x=\var{k}$ and the greatest is $x=\var{b}$ and we see that both stationary points are in $I$.

Second derivative.

The second derivative is given by:
\[\begin{eqnarray*} g''(x)&=&\simplify{4*(x-{b})^2+8*(x-{k})(x-{b})}\\ &=&\simplify{4*(x-{b})(3*x-{2*k+b})} \end{eqnarray*} \]

Local Minimum

At the stationary point $x=\var{k}$ we have $g''(\var{k})=\var{4*(k-b)^2} \gt 0$.

Hence $x=\var{k}$ is a local minimum.

The other stationary point

The value at $x=\var{b}$ is $g(\var{b})= 0$.

Hence this test fails at this point and we proceed to use the third derivative to see in more information can be gained.

Third derivative

We see that $g'''(x)=\simplify{8*(3*x-{k+2*b})}$.

Testing the stationary point using the third derivative gives:

$g'''(\var{b})=\var{8*(b-k)} \neq 0$.

Therefore there cannot be an extremum point at $x=\var{b}$.

Finding the global maximum and minimum on $I$

First we find the values at the endpoints of the interval $I=[\var{l},\var{m}]$ are:

$g(\var{l})=\var{valbegin}$.

$g(\var{m})=\var{valend}$.

Global Maximum

To find the global maximum note that we are only concerned with the values of $g$ on the interval $I$ and since $g$ does not have a local maximum on $I$ it must take its maximum value at one of the end points of $I$.

We see from the values at the end points obtained above that the global maximum value on $I$ is at $x=\var{xma}$.

We have $g(\var{xma})=\var{gma}$.

Global Minimum.

$g$ has only one local minimum on $I$ at $x=\var{k}$ and so this must be the global minimum on $I$.

We have $g(\var{k})=\var{(k-a)*(k-b)^3}$.

Question 3

Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be the function given by:
\[g(x)=\simplify{{a}*x/(x^2+{b}^2)}\]

Advice

The function $g(x)$ is continuous and differentiable at all points in $\mathbb{R}$.

Using the quotient rule for differentiation we see that
\[\begin{eqnarray*}g'(x)&=&\simplify{({a}*(x^2+{b^2})-{2*a}*x^2)/(x^2+{b^2})^2}\\ &=&\simplify{({-a}*(x-{b})(x+{b}))/(x^2+{b^2})^2} \end{eqnarray*} \]

Stationary Points.

The stationary points are given by solving $g'(x)=0$.

$g'(x)=0 \Rightarrow \simplify{{-a}*(x-{b})(x+{b})=0} \Rightarrow x=\var{b} \mbox{ or } x=\var{-b}$

The second derivative can be found by applying the quotient rule to the derivative of $g(x)$ and we obtain:

Using the quotient rule for differentiation we see that
\[\begin{eqnarray*}g''(x)&=&\simplify[std]{({-2*a}*x*(x^2+{b^2})^2+{4*a}*x*(x^2-{b^2})(x^2+{b^2}))/(x^2+{b^2})^4}\\ &=&\simplify[std]{({2*a}*x*(x^2-{3*b^2}))/(x^2+{b^2})^3} \end{eqnarray*} \]

The nature of the stationary points are determined by evaluating $g''(x)$ at the stationary points.

For $x= \var{lma}$ we have: \[g''(\var{lma})= \simplify[std]{-{abs(a)}/{2*b^3}} \lt 0\]

Hence is a local maximum.

Evaluating the function at $x=\var{lma}$ gives $g(\var{lma})=\var{valmax}$ to 3 decimal places.

For $x= \var{lmi}$ we have: \[g''(\var{lmi})= \simplify[std]{{abs(a)}/{2*b^3}} \gt 0\]

Hence is a local minimum.

Evaluating the function at $x=\var{lmi}$ gives $g(\var{lmi})=\var{valmin}$ to 3 decimal places.

The Limits.

If we divide $g(x)$ top and bottom by $x^2$ (OK as $x \neq 0$ at any time) we obtain: \[g(x)=\simplify[std]{({a}/x)/(1+{b^2}/x^2)}\]

Then using the fact that $\displaystyle \frac{1}{x}$ and $\displaystyle \frac{1}{x^2}$ both tend to $0$ as $ x \rightarrow \pm\infty$ we see that

$\displaystyle \lim_{x \to \infty}g(x)=\frac{0}{1}=0$ and similarly

$\lim_{x \to -\infty}g(x)=0$

Global Maximum and Minimum

Since $g$ has a finite limit of $0$ as $x \rightarrow \pm\infty$ and we have that $0$ lies between the local minimum $\var{valmin}$ and the local maximum $\var{valmax}$

Then:

Global Maximum: The local maximum of $g$ we have found at $x=\var{lma}$ must be a global maximum and similarly,

Global Minimum: The local minimum of $g$ we have found at $x=\var{lmi}$ must be a global minimum.

So we have shown \[\forall x \in \mathbb{R},\;\;\var{valmin} \le g(x) \le \var{valmax}\]

Question 4

Let $I=[\var{c},\var{d}]$ be an interval and let $g: I \rightarrow I$ be the function given by:
\[g(x)=\simplify{{a}*x/(x^2+{b}^2)}\]

Answer the following questions. There are seven parts and you may need to scroll down to complete all parts.

Advice

The function $g(x)$ is continuous and differentiable at all points in $\mathbb{R}$.

Stationary Points.

The stationary points are given by solving $g'(x)=0$.

$g'(x)=0 \Rightarrow \simplify{{-a}*(x-{b})(x+{b})=0} \Rightarrow x=\var{b} \mbox{ or } x=\var{-b}$

We see that both stationary points are in the inerval $I$.

The second derivative can be found by applying the quotient rule to the derivative of $g(x)$ and we obtain:

The nature of the stationary points are determined by evaluating $g''(x)$ at the stationary points.

For $x= \var{lma}$ we have: \[g''(\var{lma})= \simplify[std]{-{abs(a)}/{2*b^3}} \lt 0\]

Hence is a local maximum.

Evaluating the function at $x=\var{lma}$ gives $g(\var{lma})=\var{valmax}$ to 3 decimal places.

For $x= \var{lmi}$ we have: \[g''(\var{lmi})= \simplify[std]{{abs(a)}/{2*b^3}} \gt 0\]

Hence is a local minimum.

Evaluating the function at $x=\var{lmi}$ gives $g(\var{lmi})=\var{valmin}$ to 3 decimal places.

The values of $g$ at the endpoints are:

$g(\var{c})=\var{valbegin}$ and $g(\var{d})=\var{valend}$ to 3 decimal places.

Global Maximum and Minimum

Since $g$ has a finite limit of $0$ as $x \rightarrow \pm\infty$ and we have that $0$ lies between the local minimum value $\var{valmin}$ and the local maximum value $\var{valmax}$ (and these occur at values in $I$).

then:

Global Maximum: The local maximum of $g$ we have found at $x=\var{lma} \in I$ must be a global maximum and similarly,

Global Minimum: The local minimum of $g$ we have found at $x=\var{lmi} \in I$ must be a global minimum.

So we have shown \[\forall x \in \mathbb{R},\;\;\var{valmin} \le g(x) \le \var{valmax}\]

(all to 3 decimal places).

Question 5

Let $I=[\var{l},\var{m}]$ be an interval and let $g: I \rightarrow \mathbb{R}$ be the function given by:
\[g(x)=\simplify{x^2/(x-{c})^({a}/{b})}\]

Answer the following questions. There are seven parts and you may need to scroll down to complete all parts.

Advice

The function $g(x)$ is continuous and differentiable at all points in $\mathbb{R}$ as $ \var{c} \notin I$.

On differentiating we see that
\[\begin{eqnarray*}g'(x)&=&\simplify{2*x/(x-{c})^({a}/{b}) - {a}*x^2/({b}(x-{c})^({a+b}/{b}))}\\ &=&\simplify{({2*b}x*(x-{c})-{a}*x^2)/({b}(x-{c})^({a+b}/{b}))}\\ &=&\simplify{(x*({2*b-a}x-{2*b*c}))/({b}(x-{c})^({a+b}/{b}))} \end{eqnarray*} \]

Stationary Points.

The stationary points are given by solving $g'(x)=0$.

$\displaystyle g'(x)=0 \Rightarrow \simplify{x*({2*b-a}x-{2*b*c})=0} \Rightarrow x=0 \mbox{ or } x=\simplify[std]{{2*b*c}/{2*b-a}}$

We see that $\displaystyle x=\simplify[std]{{2*b*c}/{2*b-a}}$ is the only stationary point in $I$.

The second derivative can be found by applying the quotient rule to the derivative of $g(x)$ and we obtain:

Using the quotient rule for differentiation we see that
\[g''(x)=\simplify[std]{({a^2-3*a*b+2*b^2}*x^2+{4*b*c*(a-b)}*x+{2*c^2*b^2})/({b^2}(x-{c})^({a+2*b}/{b}))}\]

Hence \[r(x)=\simplify[std]{({a^2-3*a*b+2*b^2}*x^2+{4*b*c*(a-b)}*x+{2*c^2*b^2})}\]

The nature of the stationary points are determined by evaluating $g''(x)$ at the stationary points.

There is only one stationary point $\displaystyle x=\simplify[std]{{2*b*c}/{2*b-a}}$ in $I$ and at that point we have:
\[g''\left(\simplify[std]{{2*b*c}/{2*b-a}}\right)=\var{valsd} \gt 0\]

Hence this point is a local minimum.

Evaluating at end points of the interval.

The values of $g$ at the endpoints are:

$g(\var{l})=\var{valbegin}$ and $g(\var{m})=\var{valend}$, both to 3 decimal places.

Global Maximum and Minimum

Global Maximum: Since $g$ does not have a local maximum in the interval $I$, it must take a global maximum value at one of the end points of the interval.

From the values found for $g(\var{l})$ and $g(\var{m})$ found above, we see that $x=\var{xma}$ is the global maximum for $g$ in the interval $I$.

Global Minimum: The local minimum of $g$ given by $ \displaystyle x=\simplify[std]{{2*b*c}/{2*b-a}} \in I$ is the only local minimum and must be a global minimum in $I$.

Note that the global minimum value for $g$ on $I$ is:

\[g\left(\simplify[std]{{2*b*c}/{2*b-a}}\right)=\var{valmin}\]

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Maths Support: Maxima and minima for differentiable functions on intervals

Question 1

Question 1

a)

b)

Find the stationary points of $g$.

Gap 2

c)

d)

e)

Global Maximum

Global Minimum

Advice

Local Maximum

Local Minimum

Global Maximum

Global Minimum

Question 2

Question 2

a)

b)

Find the stationary points of $g$, here thought of as $g:\mathbb{R} \rightarrow \mathbb{R}$.

Gap 2

c)

Gap 1

Gap 2

d)

Gap 3

e)

f)

Global Maximum

Global Minimum

Advice

First derivative

Stationary points

Second derivative.

Local Minimum

The other stationary point

Third derivative

Finding the global maximum and minimum on $I$

Global Maximum

Global Minimum.

Question 3

Question 3

a)

b)

c)

d)

Find the stationary points of $g$.

e)

f)

Gap 0

Gap 1

Gap 2

Gap 2

Advice

Stationary Points.

The Limits.

Global Maximum and Minimum

Question 4

Question 4

a)

b)

c)

d)

Find the stationary points of $g$.

Gap 2

e)

f)

g)

Global Maximum

Global Minimum

Advice

Stationary Points.

Global Maximum and Minimum

Question 5

Question 5

a)

b)