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Tutorial 12

1 .The function is decreasing only on the interval (0,3).Find

Since f(x) is only decreasing over interval (0,3), are critical value.

, since a is a constant.

2. The function has a relative minimum at x = 2.Find a. At

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3. Find the point on the parabola nearest to the point (-3,0).The possible value of x nearer to is

When

Find distance between (-3,0) and (-1,1)

When Find distance between (-3,0) and (-2,4)

When Find distance between (-3,0) and (-3,9)

The nearest distance is 2.236

The point in parabola nearer to point (-3,0) is (-1,1)

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4. The sum of two positive numbers is 50. If x denotes one of these numbers, for which number x is

the product of the two numbers an increasing function of x, what is the maximum value of their

product?

ANSWER:

Assume 1st

number is x and 2nd

number is y

Thus,

.. (1)

Since x is the product of numbers in a function,

.(2)

Substitute (1) in (2)

Then, differentiate the equation

Substitute x=25 into x+y=50

Thus, maximum value of their product is

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5. Let

a) Find the intervals on which is an increasing function.b) Use the result of part (a) to conclude that for all

Sol:

(a)

Interval

Conclusion

+ Increasing - Decreasing - Decreasing

+ Increasing

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(b)

Tested value of at interval

(c)

Sol:

Find :-

Substitute into-

The point is relative minimum since

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(d) Sol:

Find-

Substitute into:-

Point is a relative maximum since .

6. Find the interval on which is increasing or decreasing. Find all local extrema fora)

When

When

Thus, stationary point is

Interval Sign of Decreasing Increasing

Since

changes sign negative to positive at

, point

is minimum

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b) When

At At

Interval Sign of Increasing

Decreasing Increasing

Since changes sign positive to negative, point is maximum point.Since changes sign negative to positive, point is minimum point.

(c)

Solution :

=0

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=0

3

Interval Conclusion(- ) decrease(0 ) increase

When Thus point (0,4) is minimum point

d)

Solution :

3

Interval Conclusion(- ) Increase(0,) Increase

As the interval of shows no different, no local extrema is plotted

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(e)

Find critical point :

At ,

Interval Test point Conclusion + is increasing. - is decreasing.

Point

is a

relative maximum.

Function is increasing in interval , and decreasing in interval .

Find maximum point :

At ,

The maximum point is at .

Since has a modulus,.Hence, minimum point(s) is exist at .

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Find minimum point :

At ,

The minimum point is at .

(f) .

Find critical point :

At , . /

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Interval Test point Conclusion

+ is increasing.

-

is decreasing.

Point is arelative maximum.

+ is increasing.Point is arelative minimum.

Function is increasing in interval . / and , while decreasing in interval

.

Find maximum point :

At

The maximum point is at .

Find minimum point :

At ,

The minimum point is at

.

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(g)

, ,

Interval Decrease Increase

(h)

Interval

Decrease Decrease

,

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i) {

0 1

Sign f(x) +ve -ve +ve

The relative minimum value , Minimum point is (0,4)The relative maximum value Maximum point is (1,3)

j)

sign f(x) +ve +ve +ve

No local extremum

k)

Let

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Interval Sign of

Increase Increase

Since,does not changes its sign, there is no local extremum

l)

Let

Interval

Sign of Decrease Increase

There is a relative minimum at

When

is a minimum point

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m)

0 2

Sign f(x) +ve -ve +ve

F(x) increasing decreasing increasing

The relative maximum value is (0,3)

The relative minimum value is (0, )

n)

0 2 4

Sign f(x) -ve -ve +ve +ve

The minimum point is The coordinates is (2,-64)

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Tutorial 13

1. Determine the interval s on which the graph of is concave up or concave down. Find anyinflection points.

a)

Thus, the function is concave up on and there is no inflection point.

b)

To get intervals, let

Interval Sign of

Concavity

. / Concave downward

Concave upward

Concave upward

There are points of inflection at and

When

When

Hence, the points of inflection are . / and . /

(c)

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Interval Increase Increase Concave

down

Concave up

(d)

Interval No value Increase No value Concave

down

(e)

Find critical point :

not exists at .

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Interval Test point

Conclusion

+ isconcave up.

- isconcave down.

Since not exists at

,point is an

asymptote. No

inflection point

is exists.

Function is concave up in interval , while concave down in interval .

(f)

Find critical point :

At ,

Interval Test point

Conclusion

- is

concave down.

+ is

concave up.

Point isan inflection

point.

Function is concave up in interval

while concave down in interval .

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Find inflection point :

At , , -

Inflection point is at .

g) Solution :

Interval Conclusion(-) Concave upward(-1,0) Concave downward(0,1) Concave downward(1, ) Concave downwardWhen Inflection point at

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h) Solution :

Interval Conclusion() Concave downward() Concave downward(0,1) Concave upward(1,) Concave upward

When Inflection point at (0,3)

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2. Determine whether has a relative extremum at the given value of, using either First orSecond Derivative Test.

a)

At ./

relative maximum

b)

Intervals Sign of

Since changes sign negative to positive, is minimum at

(c)

Sol:

Find :-

Substitute into- The point is relative minimum since

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(d) Sol:

Find- Substitute into:- Point is a relative maximum since .

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Tutorial 14

Sketch the graphs of the following functions:

1) ANSWER:Differentiate the function

Find the value of x, since f=0

Use interval table

0 +ve -ve +ve

-ve +ve +ve

Find the x-intercept, as y=0

Find the y-intercept, as x=0

The graph of

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2) ANSWER:

Differentiate the function

Find the value of x, since f=0

Since f is undefined, thus

Use table

+ve +ve

+ve -ve

Find the x-intercept, as y=0

Find the value of y, as x=0

=undefined

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The graph of

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3.

The domain offisx0

When When

The stationary point is (1,3)

Interval f(x) Conclusion +ve fis increasing (1,) -ve fis decreasing (-,1) but doesnt touch

*+Then we findf(x) to determine the concavity,

Interval f(x) Conclusion

-ve f(x) is concave down at ( except at x=0

+ve f(x) is concave up at

Thus, is the inflection point since that point become the centre between concave updown and concave down

Find horizontal asymptote:-

=0

Find horizontal asymptote:-

Use limit to infinity,

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4. The domain offis x0, 4

use quotient rule

when x=2, y=-0.25Interval f(x) is increasing in interval is decreasing in interval and There is a relative maximum at point Findf(x) to determine concavity of function.

quotient rule

, -

Interval f(x) is concave up in interval is concave down in interval and Find horizontal asymptote:-

Use limit to infinity,

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5.

at ,

Interval

Conclusion Concavity

Decreasing Concavedown

Increasing Concavedown

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6.

at therefore,

then find the point of inflection,

Since the signs of changes from to , it proves that -1 is the inflection point.

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7) Use quotient rule

()

:Interval

Conclusion

Use quotient rule

()

,()-

Interval Conclusion

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8) Use quotient rule

()

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:Interval Conclusion

Use quotient rule

()

,()-

Interval Conclusion