Math Editor

Assignment: MTH100

NAME: Ayesha Arshad

ID: bc260205730

Question no :1

Give two functions f(x)=4x+2 and g(x)=2x+3. Find the composite functions (f.g)(x) and (g.f)(x).

Solution:

f(x)=4x+2g(x)=2x+3Firatst case:(f.g)(x)=f(g(x))=f(2x+3)=4(2x+3)+2=8x+12+2=8x+14Second case:(g.f)(x)=g(f(x))=g(4x+2)=2(4x+2)+3=8x+4+3=8x+7so thw the answers for the two given functions are :(f.g)(x)=8x+14(g.f)(x)=8x+7​

Question no 2:

given f(x)=7y+13. Find inverse of it.

Solution:

​f(x)=7y+13​

y=f(x)

Swapping x with y

x=7y3+13

Solving for "y"

x=13−7y3

• Taking cuberoot on bth side

³√√x−13​=³√7y3​

7y=³√x−13​

y= 3√7x−13​​

Question no 3:

Determine weather f(x)=√x3+7​ is even or odd. Also find it's range and domain.

Solution:

a. Finding even or odd:

Tob find either the given function is even or odd first we have to find f(−x) of given function so,

f(−x)=√(−x3)+7​

f(−x)=√−x3+7​

As we know for even,

f(−x)=f(x)​

And for odd'

f(−x)=−f(x)​

"So it shows that the given function is neither even nor odd."

b. Domain:

Taking the inner values of square root and thet must be non negative

x3+7≥0​

x3≥0−7x3≥−7taking cube root on both sides3√x3≥3√−7​x≥3√−7​so the domain will be[(3√−7​) ,∞]]Range:for the smallest range of xwell we’ll put:f(−3√−7​)= √0=0for the largest value of x:3x3+7=∞√x3+7​=∞so the range will be:(0,∞)​

Question no 4:

Evaluate:

∣∣∣∣​6+8i7+13i​∣∣∣∣​

Solution:

To solve some complex numbers, we use a method

∣6+8i∣∣7+13i∣​We have to solve themm separately as numerator and denominator. We'll be using the given formula: ∣a+bi∣=√a2+b2​

For numerator:

∣7+13∣=√(7)2+(13)2​​​

=√49+169​

=√218

For denominator:

∣6+8i∣​

=√(6)2+(8)2​

=√36+64​

=√100

=10so for ratio of numerator n denominator :=10√218​ ​

This is the possible lowest form.

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