Question

Find x(t). if X(s) = 1/5^2+4s+8

Answer 1

Answer:

Domain : R−{−1,3}
Range: R

Explanation:

To Find the domain

Equate the denominator(x2−2x−3) to zero, then solve the equation for x

→x2−2x−3=0
→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1
→x=1±2

⇒x=−1 and x=3

This means that, when x=−1 or 3, we have the x2−2x−3=0
Implying that f(x)=x0 which is undefined.

Hence, the domain is all real numbers except −1
and 3

Also written as Df=(−∞,−1)∪(−1,3)∪(3,+∞)

To Find the Range

Step 1
say f(x)=y and rearrange the function as a quadratic equation

y=xx2−2x−3
→y(x2−2x−3)=x
→yx2−2yx−3y−x=0
→yx2+(−2y−1)x−3y=0

Step 2
We know from the quadratic formula,
x=−b±√b2−4ac2a
that the solutions of x are real when b2−4ac≥0

So likewise, we say, (−2y−1)2−4⋅(y)⋅(−3y)≥0
b a c

Step 3
We solve the inequality for the values set of values of y

⇒4y2+4y+1+12y2≥0

→16y2+4y+1≥0

→16[y2+14y+116]≥0

→16[(y+18)2−164+116]≥0

→16[(y+18)2+364]≥0

Notice that for all values of y the left hand side of the inequality be greater than (but not equal) to zero.

We then conclude that, y can take all real values.

y∈R⇔f(x)∈R

So the Range is R