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MTH101 Calculus And Analytical Geometry Assignment No 1 Solution & Discussion Due Date:12-11-2012

Assignment No. 1

MTH 101 (Fall 2012)

Maximum Marks: 15

Due Date: 12th Nov. 2012

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Question # 01:                                                                                        05

Find the equation for circle with center (1,3) that passes through (4, -1).

Question # 02:                                                                                        05

Given that . Find . Also find the domains of the composite functions.

Question # 03:                                                                                        05

Find . (Use rationalization technique to simplify)

Hint: (How to rationalize)

Rationalize the numerator of the expression .

Step I: find the conjugate of the numerator which is and vice versa.

Step II: multiply  the numerator and the denominator of the expression with the conjugate

Step III: make sure all the radicals are simplified

Step IV: simplify the fraction if needed.

In the same manner we can rationalize denominator too.

(Note: In order to get full marks, do all necessary steps)

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Replies to This Discussion

All the values that go into a function

The output values are called the range.

Domain -> function -> Range

Example: if the function f(x) = x2 is given the values x = {1,2,3,...} then {1,2,3,...} is the domain.

Question # 01:                                                                                        05

Find the equation for circle with center (1,3) that passes through (4, -1).

Solution:-

Radius of the circle r is the distance between (4, -1) & (3, 1)

r =

r =

r =

r =

r =        5

Now we put the value of r in equation of circle

(x – 1)2 + (y – 3)2 + 25

x2 +1 – 2x + y2 + 9 – 6y = 25

x2+ y – 2x – 6y + 10 = 25

x2+ y – 2x – 6y - 15 = 0

Question # 02:                                                                                        05

Given that . Find . Also find the domains of the composite functions.

Solution:-

= (g(x))2 – 4

=

=

=

The Domain of   all Real Number

Now   =

=

=

=

The Domain of

Taking under root on both sides of inequality to eliminate the exponent

First substitute the  portion of the  to find the solution

Now substitute the  portion of the to find the solution

The Domain of  =  is     or

Question # 03:                                                                                        05

Find . (Use rationalization technique to simplify)

Solution:-

X

1

As  approaches to , then  also approaches to

The of  is

=    0

Question # 01:                                                                                        05

Find the equation for circle with center (1,3) that passes through (4, -1).

Solution:-

Radius of the circle r is the distance between (4, -1) & (3, 1)

r =

r =

r =

r =

r =        5

Now we put the value of r in equation of circle

(x – 1)2 + (y – 3)2 + 25

x2 +1 – 2x + y2 + 9 – 6y = 25

x2+ y – 2x – 6y + 10 = 25

x2+ y – 2x – 6y - 15 = 0

Question # 02:                                                                                        05

Given that . Find . Also find the domains of the composite functions.

Solution:-

= (g(x))2 – 4

=

=

=

The Domain of   all Real Number

Now   =

=

=

=

The Domain of

Taking under root on both sides of inequality to eliminate the exponent

First substitute the  portion of the  to find the solution

Now substitute the  portion of the to find the solution

The Domain of  =  is     or

Question # 03:                                                                                        05

Find . (Use rationalization technique to simplify)

Solution:-

X

1

As  approaches to , then  also approaches to

The of  is

=    0

fog(x)
f(g(x))
f(underroot x-2)
f(underroot (x-2)^2 -4)
f(x-2-4)
fog(x)=x-6
what is domain of it

where:
f(x) = (3 - x)^1/2
g(x)= (x^2 - 16)^1/2
The above answer is incorrect... where is why, when he writes:
-----------------
f o g(x)
= f[g(x)]
= f[√(x^2 - 16)]
= √[3 - (x^2 - 16)]
= √(3 - x^2 + 16)
= √(-x^2 + 19)

when he subs plugs it into f( ), he forgets the square root sign which changes his answer.

Also,

in the next part,

g o f(x) =
= g[f(x)]
= g[√(3 - x)]
= √[(√3 - x)^2 - 16)
= √(3 - 2x√3 + x^2 - 16)
= √(x^2 - 2x√3 - 13)

when he plugs into g( ), he should be plugging in sqrt(3-x) instead he plugs in sqrt(3) - x, which leads to the wrong answer.
-----------

I've copy/pasted the correct solution which I wrote on your other question,

-------
you typed f(x) and g(x) perfectly. sqrt(a) = a^1/2 = square root of a, are all acceptable.

by the way i use the notation "s.t." to mean "such that"

Ok so basically the domain is the "allowable x values"

so for example in this case, we know you cant take a square root of a negative number so you need to restrict the possible x values so that it never happens.

so for f(x) = (3 - x )^1/2

we want to make sure that 3 - x >= 0 (so that we can take the square root,

which means 3 >= x or which u can write as x <= 3.

so formally, domain( f ) = {x s.t. x <= 3}

for g(x), its the same basic argument, we need to make sure that inside the square root is always >= 0, so...

x^2 - 16 >= 0
x^2 >= 16
x >= +/- root(16)
x >= +/- 4
which means either x >= +4 or x <= -4 (since when you divide/mulitply by a negative number you need to flip the inequality sign)

so the domain( g ) = {x s.t. x >= 4 or x<= -4}

Now to the fog(x) and gof(x),

these are basically f( g(x) ) and g( f(x) ).

So basically you evaluate g(x) and plug that in for an input to f(x), and vice versa.

fog(x) = f( g(x )) = f( (x^2 - 16)^1/2 ) = (3 - [(x^2 - 16)^1/2])^1/2

again we must make sure that we can take the square root,

so we need 3 - [(x^2 - 16)^1/2] >= 0

meaning -(x^2 - 16) >= -9

or x^2 =< 25
or x =< +/-5
meaning x <= 5 AND x >= -5

so, domain((fog)(x) ) = {x : -5 <= x <= 5 }

similarly,

gof(x) = g( f(x) ) = ([(3 - x)^1/2]^2 - 16)^1/2 = ((3 - x) - 16)^1/2
gof(x) = (-x -13)^1/2

so we need -x - 13 >= 0, meaning -x >= 13 or x <= -13.

so domain((gof)(x) ) = {x : x <= -13}

As an aside, here we were restricting the domain based on making sure a squareroot was positive... another very common form is making sure you dont divide by 0.

so for example,

h(x ) = 10/ (x - 5)

our domain would be all x s.t. x =/= 5 (x not equal to 5)

and you can also have combinations of both for example

t(x) = 10 / sqrt( x - 5)

so here we need that x - 5 > 0 and x =/= 5, we can solve for both at the same time by just saying we need sqrt( x - 5 ) > 0. since if it = 0 then we divide by 0.

so, sqrt( x - 5) > 0,
x - 5 > 0
x > 5

so here your domain would be x s.t. x > 5

What is the domain of the Q.2?

the ans of fog(x) and gof(x) should be equal?

in solution 3 if we open the formula a2-b2 then it will ne (a+b)(a-b)

question No 2

the domain of 1 question is (- infinity , + infinity) all real numbers ,

the domain of [6, -infinity)

the composite domain is  [6, -infinity)

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