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
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√−7x≥3√−7so 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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