Determine the relative maximum and minimum on the graph

In summary, we discussed the given function and its derivative graph, and answered various questions related to critical points, intervals of increasing and decreasing functions, and absolute maximum and minimum points on the graph. We determined that the critical points are 4.5 and 13, and the intervals of increasing and decreasing functions are [0,4.5], (10,13], [4.5,10), and [13, -∞). The absolute maximum point is 4.5, and the absolute minimum point is 10. We also discussed the concept of relative maxima and minima and how they can be identified based on the value of the derivative at certain points.
  • #1
58
0
Given that f is the function on (−∞, ∞) and the graph is the derivative of f

1.) Find the critical point on the graph ?
2.) Find the interval of the increasing function on the graph ?
3.) Find the interval of the decreasing function on the graph ?
4.) Find the point which is the absolute maximum on the graph ?
5.) Find the point which is the absolute minimum on the graph ?
func.png
 
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  • #2
jaychay said:
Given that f is the function on (−∞, ∞) and the graph is the derivative of f

1.) Find the critical point on the graph ?
2.) Find the interval of the increasing function on the graph ?
3.) Find the interval of the decreasing function on the graph ?
4.) Find the point which is the absolute maximum on the graph ?
5.) Find the point which is the absolute minimum on the graph ?View attachment 10704
Do the graph have the relative maximum and relative minimum ?
 
  • #3
Let's add the graph of $\color{red}f(x)$ at an arbitrary level. That is, let's pick $\color{red}f(0)=0$.
And let's assume that $\color{red}f(x)$ is continuous at $x=10$.

\begin{tikzpicture}[
declare function={
df1(\x) = 2*cos(3/11*360)-2*cos((\x+1)/11*360);
df2(\x) = 1.25-5/16*(\x-13)^2;
f1(\x) = 2*cos(3/11*360)*\x-2*(sin((\x+1)/11*360) - sin(1/11*360))*11/(2*pi);
f2(\x) = 1.25*(\x-10)-5/16*((\x-13)^3 + 27)/3+f1(10);
}]
%\draw[help lines] (-1,-3) grid (16,4);
\draw[-latex] (-1,0) -- (16,0);
\draw[-latex] (0,-3) -- (0,4);
\draw foreach \i in {1,...,15} { (\i,0.1) -- (\i,-0.1) node[below] {$\i$} };
% \draw foreach \i in {-2,...,2} { (0.1,\i) -- (-0.1,\i) node[ left ] {$\i$} };
\draw[domain=-1:10, variable=\x, thick, smooth] plot ({\x}, {df1(\x)}) (3, {df1(3)}) node[above left] {$f'(x)$};
\draw[domain=10:16, variable=\x, thick, smooth] plot ({\x}, {df2(\x)});
\filldraw[fill=black!5, thick] (10,{df1(10)}) circle (0.05) (10,{df2(10)}) circle (0.05);
\draw[domain=-1:10, variable=\x, red, thick, smooth] plot ({\x}, {f1(\x)}) (9, {f1(9)}) node[above right] {$f(x)$};;
\draw[domain=10:16, variable=\x, red, thick, smooth] plot ({\x}, {f2(\x)});
\end{tikzpicture}

Can we find those points and intervals now?
 
  • #4
Klaas van Aarsen said:
Let's add the graph of $\color{red}f(x)$ at an arbitrary level. That is, let's pick $\color{red}f(0)=0$.
And let's assume that $\color{red}f(x)$ is continuous at $x=10$.

\begin{tikzpicture}[
declare function={
df1(\x) = 2*cos(3/11*360)-2*cos((\x+1)/11*360);
df2(\x) = 1.25-5/16*(\x-13)^2;
f1(\x) = 2*cos(3/11*360)*\x-2*(sin((\x+1)/11*360) - sin(1/11*360))*11/(2*pi);
f2(\x) = 1.25*(\x-10)-5/16*((\x-13)^3 + 27)/3+f1(10);
}]
%\draw[help lines] (-1,-3) grid (16,4);
\draw[-latex] (-1,0) -- (16,0);
\draw[-latex] (0,-3) -- (0,4);
\draw foreach \i in {1,...,15} { (\i,0.1) -- (\i,-0.1) node[below] {$\i$} };
% \draw foreach \i in {-2,...,2} { (0.1,\i) -- (-0.1,\i) node[ left ] {$\i$} };
\draw[domain=-1:10, variable=\x, thick, smooth] plot ({\x}, {df1(\x)}) (3, {df1(3)}) node[above left] {$f'(x)$};
\draw[domain=10:16, variable=\x, thick, smooth] plot ({\x}, {df2(\x)});
\filldraw[fill=black!5, thick] (10,{df1(10)}) circle (0.05) (10,{df2(10)}) circle (0.05);
\draw[domain=-1:10, variable=\x, red, thick, smooth] plot ({\x}, {f1(\x)}) (9, {f1(9)}) node[above right] {$f(x)$};;
\draw[domain=10:16, variable=\x, red, thick, smooth] plot ({\x}, {f2(\x)});
\end{tikzpicture}

Can we find those points and intervals now?
My answers for all of the question that I have done on my own is
  • 1.) 4.5,13
  • 2.) [0,4.5] , (10,13]
  • 3.) [4.5,10), [13, - ∞)
  • 4.) 4.5,13
  • 5.) 10
I want to check that I am doing it correct or not because I am not good at discontinuous graph
Can you please help me
Thank you in advice
 
  • #5
jaychay said:
Given that f is the function on (−∞, ∞) and the graph is the derivative of f

1.) Find the critical point on the graph ?
  • 1.) 4.5,13

A critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero.

That's not the case for x=4.5 or x=13 is it?
For which x-values is $f'(x)$ either zero or undefined?

jaychay said:
2.) Find the interval of the increasing function on the graph ?
3.) Find the interval of the decreasing function on the graph ?
  • 2.) [0,4.5] , (10,13]
  • 3.) [4.5,10), [13, - ∞)

A function is increasing if its derivative is greater than zero.
Where is $f'(x)$ greater than zero?

We can also look at the graph for $f(x)$ that I added.
Where is that graph increasing?

jaychay said:
4.) Find the point which is the absolute maximum on the graph ?
5.) Find the point which is the absolute minimum on the graph ?
  • 4.) 4.5,13
  • 5.) 10

We have relative maxima and minima if $f'(x)=0$.
For which x-values is $f'(x)$ equal to zero?
Can we tell whether they correspond to maxima or minima?
 
  • #6
Klaas van Aarsen said:
A critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero.

That's not the case for x=4.5 or x=13 is it?
For which x-values is $f'(x)$ either zero or undefined?
A function is increasing if its derivative is greater than zero.
Where is $f'(x)$ greater than zero?

We can also look at the graph for $f(x)$ that I added.
Where is that graph increasing?
We have relative maxima and minima if $f'(x)=0$.
For which x-values is $f'(x)$ equal to zero?
Can we tell whether they correspond to maxima or minima?
Thank you very much
 

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