The extreme value theorem states that a continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. As shown in
[link] , one or both of these absolute extrema could occur at an endpoint. If an absolute extremum does not occur at an endpoint, however, it must occur at an interior point, in which case the absolute extremum is a local extremum. Therefore, by
[link] , the point
at which the local extremum occurs must be a critical point. We summarize this result in the following theorem.
Location of absolute extrema
Let
be a continuous function over a closed, bounded interval
The absolute maximum of
over
and the absolute minimum of
over
must occur at endpoints of
or at critical points of
in
With this idea in mind, let’s examine a procedure for locating absolute extrema.
Problem-solving strategy: locating absolute extrema over a closed interval
Consider a continuous function
defined over the closed interval
Evaluate
at the endpoints
and
Find all critical points of
that lie over the interval
and evaluate
at those critical points.
Compare all values found in (1) and (2). From
[link] , the absolute extrema must occur at endpoints or critical points. Therefore, the largest of these values is the absolute maximum of
The smallest of these values is the absolute minimum of
Now let’s look at how to use this strategy to find the absolute maximum and absolute minimum values for continuous functions.
Locating absolute extrema
For each of the following functions, find the absolute maximum and absolute minimum over the specified interval and state where those values occur.
over
over
Step 1. Evaluate
at the endpoints
and
Step 2. Since
is defined for all real numbers
Therefore, there are no critical points where the derivative is undefined. It remains to check where
Since
at
and
is in the interval
is a candidate for an absolute extremum of
over
We evaluate
and find
Step 3. We set up the following table to compare the values found in steps 1 and 2.
Conclusion
Absolute maximum
Absolute minimum
From the table, we find that the absolute maximum of
over the interval [1, 3] is
and it occurs at
The absolute minimum of
over the interval [1, 3] is
and it occurs at
as shown in the following graph.
Step 1. Evaluate
at the endpoints
and
Step 2. The derivative of
is given by
for
The derivative is zero when
which implies
The derivative is undefined at
Therefore, the critical points of
are
The point
is an endpoint, so we already evaluated
in step 1. The point
is not in the interval of interest, so we need only evaluate
We find that
Step 3. We compare the values found in steps 1 and 2, in the following table.
Conclusion
Absolute maximum
Absolute minimum
We conclude that the absolute maximum of
over the interval [0, 2] is zero, and it occurs at
The absolute minimum is −2, and it occurs at
as shown in the following graph.
Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
No. According to Isac Newtons law. this two bodies maybe you and the wall beside you.
Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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