5.1 Differentiation (first principles, rules) and sketching graphs  (Page 3/3)

 Page 3 / 3
$\begin{array}{ccc}\hfill 2x& =& 0\hfill \\ \hfill x& =& 0\hfill \end{array}$

We then substitute this into the equation of the graph (i.e. $y={x}^{2}$ ) to determine the $y$ -coordinate of the turning point: $f\left(0\right)={\left(0\right)}^{2}=0$ This corresponds to the point that we have previously calculated.

Calculate the turning points of the graph of the function $f\left(x\right)=2{x}^{3}-9{x}^{2}+12x-15$ .

1. Using the rules of differentiation we get: ${f}^{\text{'}}\left(x\right)=6{x}^{2}-18x+12$

2. $\begin{array}{ccc}\hfill 6{x}^{2}-18x+12& =& 0\hfill \\ \hfill {x}^{2}-3x+2& =& 0\hfill \\ \hfill \left(x-2\right)\left(x-1\right)& =& 0\hfill \end{array}$

Therefore, the turning points are at $x=2$ and $x=1$ .

3. $\begin{array}{ccc}\hfill f\left(2\right)& =& 2{\left(2\right)}^{3}-9{\left(2\right)}^{2}+12\left(2\right)-15\hfill \\ & =& 16-36+24-15\hfill \\ & =& -11\hfill \end{array}$
$\begin{array}{ccc}\hfill f\left(1\right)& =& 2{\left(1\right)}^{3}-9{\left(1\right)}^{2}+12\left(1\right)-15\hfill \\ & =& 2-9+12-15\hfill \\ & =& -10\hfill \end{array}$
4. The turning points of the graph of $f\left(x\right)=2{x}^{3}-9{x}^{2}+12x-15$ are (2,-11) and (1,-10).

We are now ready to sketch graphs of functions.

Method:

Sketching Graphs: Suppose we are given that $f\left(x\right)=a{x}^{3}+b{x}^{2}+cx+d$ , then there are five steps to be followed to sketch the graph of the function:

1. If $a>0$ , then the graph is increasing from left to right, and may have a maximum and a minimum. As $x$ increases, so does $f\left(x\right)$ . If $a<0$ , then the graph decreasing is from left to right, and has first a minimum and then a maximum. $f\left(x\right)$ decreases as $x$ increases.
2. Determine the value of the $y$ -intercept by substituting $x=0$ into $f\left(x\right)$
3. Determine the $x$ -intercepts by factorising $a{x}^{3}+b{x}^{2}+cx+d=0$ and solving for $x$ . First try to eliminate constant common factors, and to group like terms together so that the expression is expressed as economically as possible. Use the factor theorem if necessary.
4. Find the turning points of the function by working out the derivative $\frac{df}{dx}$ and setting it to zero, and solving for $x$ .
5. Determine the $y$ -coordinates of the turning points by substituting the $x$ values obtained in the previous step, into the expression for $f\left(x\right)$ .
6. Use the information you're given to plot the points and get a rough idea of the gradients between points. Then fill in the missing parts of the function in a smooth, continuous curve.

Draw the graph of $g\left(x\right)={x}^{2}-x+2$

1. The $y$ -intercept is obtained by setting $x=0$ . $g\left(0\right)={\left(0\right)}^{2}-0+2=2$ The turning point is at (0,2).

2. The $x$ -intercepts are found by setting $g\left(x\right)=0$ .

$\begin{array}{ccc}\hfill g\left(x\right)& =& {x}^{2}-x+2\hfill \\ \hfill 0& =& {x}^{2}-x+2\hfill \end{array}$

Using the quadratic formula and looking at ${b}^{2}-4ac$ we can see that this would be negative and so this function does not have real roots. Therefore, the graph of $g\left(x\right)$ does not have any $x$ -intercepts.

3. Work out the derivative $\frac{dg}{dx}$ and set it to zero to for the $x$ coordinate of the turning point. $\frac{dg}{dx}=2x-1$

$\begin{array}{ccc}\hfill \frac{dg}{dx}& =& 0\hfill \\ \hfill 2x-1& =& 0\hfill \\ \hfill 2x& =& 1\hfill \\ \hfill x& =& \frac{1}{2}\hfill \end{array}$
4. $y$ coordinate of turning point is given by calculating $g\left(\frac{1}{2}\right)$ .

$\begin{array}{ccc}\hfill g\left(\frac{1}{2}\right)& =& {\left(\frac{1}{2}\right)}^{2}-\left(\frac{1}{2}\right)+2\hfill \\ & =& \frac{1}{4}-\frac{1}{2}+2\hfill \\ & =& \frac{7}{4}\hfill \end{array}$

The turning point is at $\left(\frac{1}{2},\frac{7}{4}\right)$

Sketch the graph of $g\left(x\right)=-{x}^{3}+6{x}^{2}-9x+4$ .

1. Find the turning points by setting ${g}^{\text{'}}\left(x\right)=0$ .

If we use the rules of differentiation we get ${g}^{\text{'}}\left(x\right)=-3{x}^{2}+12x-9$

$\begin{array}{ccc}\hfill {g}^{\text{'}}\left(x\right)& =& 0\hfill \\ \hfill -3{x}^{2}+12x-9& =& 0\hfill \\ \hfill {x}^{2}-4x+3& =& 0\hfill \\ \hfill \left(x-3\right)\left(x-1\right)& =& 0\hfill \end{array}$

The $x$ -coordinates of the turning points are: $x=1$ and $x=3$ .

The $y$ -coordinates of the turning points are calculated as:

$\begin{array}{ccc}\hfill g\left(x\right)& =& -{x}^{3}+6{x}^{2}-9x+4\hfill \\ \hfill g\left(1\right)& =& -{\left(1\right)}^{3}+6{\left(1\right)}^{2}-9\left(1\right)+4\hfill \\ & =& -1+6-9+4\hfill \\ & =& 0\hfill \end{array}$
$\begin{array}{ccc}\hfill g\left(x\right)& =& -{x}^{3}+6{x}^{2}-9x+4\hfill \\ \hfill g\left(3\right)& =& -{\left(3\right)}^{3}+6{\left(3\right)}^{2}-9\left(3\right)+4\hfill \\ & =& -27+54-27+4\hfill \\ & =& 4\hfill \end{array}$

Therefore the turning points are: $\left(1,0\right)$ and $\left(3,4\right)$ .

2. We find the $y$ -intercepts by finding the value for $g\left(0\right)$ .

$\begin{array}{ccc}\hfill g\left(x\right)& =& -{x}^{3}+6{x}^{2}-9x+4\hfill \\ \hfill {y}_{int}=g\left(0\right)& =& -{\left(0\right)}^{3}+6{\left(0\right)}^{2}-9\left(0\right)+4\hfill \\ & =& 4\hfill \end{array}$
3. We find the $x$ -intercepts by finding the points for which the function $g\left(x\right)=0$ .

$g\left(x\right)=-{x}^{3}+6{x}^{2}-9x+4$

Use the factor theorem to confirm that $\left(x-1\right)$ is a factor. If $g\left(1\right)=0$ , then $\left(x-1\right)$ is a factor.

$\begin{array}{ccc}\hfill g\left(x\right)& =& -{x}^{3}+6{x}^{2}-9x+4\hfill \\ \hfill g\left(1\right)& =& -{\left(1\right)}^{3}+6{\left(1\right)}^{2}-9\left(1\right)+4\hfill \\ & =& -1+6-9+4\hfill \\ & =& 0\hfill \end{array}$

Therefore, $\left(x-1\right)$ is a factor.

If we divide $g\left(x\right)$ by $\left(x-1\right)$ we are left with: $-{x}^{2}+5x-4$ This has factors $-\left(x-4\right)\left(x-1\right)$

Therefore: $g\left(x\right)=-\left(x-1\right)\left(x-1\right)\left(x-4\right)$

The $x$ -intercepts are: ${x}_{int}=1,4$

Sketching graphs

1. Given $f\left(x\right)={x}^{3}+{x}^{2}-5x+3$ :
1. Show that $\left(x-1\right)$ is a factor of $f\left(x\right)$ and hence fatorise $f\left(x\right)$ fully.
2. Find the coordinates of the intercepts with the axes and the turning points and sketch the graph
2. Sketch the graph of $f\left(x\right)={x}^{3}-4{x}^{2}-11x+30$ showing all the relative turning points and intercepts with the axes.
1. Sketch the graph of $f\left(x\right)={x}^{3}-9{x}^{2}+24x-20$ , showing all intercepts with the axes and turning points.
2. Find the equation of the tangent to $f\left(x\right)$ at $x=4$ .

Local minimum, local maximum and point of inflextion

If the derivative ( $\frac{dy}{dx}$ ) is zero at a point, the gradient of the tangent at that point is zero. It means that a turning point occurs as seen in the previous example.

From the drawing the point (1;0) represents a local minimum and the point (3;4) the local maximum .

A graph has a horizontal point of inflexion where the derivative is zero but the sign of the sign of the gradient does not change. That means the graph always increases or always decreases.

From this drawing, the point (3;1) is a horizontal point of inflexion, because the sign of the derivative does not change from positive to negative.

where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Got questions? Join the online conversation and get instant answers!