**Question 1**

Sketch f(x) = x5 −3×4 −x3 + 7×2 −4. without an graphing assistance

Question 2

If g ( x ) = x 2 + 4 x with x ≥ − 2 , find g − 1 ( 5 ) . Inverse Function: In order to solve the inverse of a function, we need to use some algebraic techniques. On the other hand, we also need to interchange x and y and solve for y.

Question 3

Solve the equation.

4×2−49=04×2−49=0

A quadratic equation in a standard form:

ax2+bx−c=0ax2+bx−c=0

Can be solved using a quadratic formula. Suppose, however, that b=0b=0 to produce a special case in which ax2−c=0ax2−c=0.

This equation can be solved using the factorization formula:

ax2−c=(√ax−√c)(√ax+√c)

Question 4

For f(x)=4×2+3x−7f(x)=4×2+3x−7 and g(x)=x−1g(x)=x−1,

a) Find (f/g)(x)(f/g)(x) and simplify.

b) Determine the domain of (f/g)(x)(f/g)(x). Write your answer in interval notation.

Operation on functions means two functions with overlapping values of the domain can be added, subtracted, multiplied, and divided together. For example, given the functions f(x)f(x) and g(x)g(x) for all values of the variable in the domain of both functions, we can perform the four basic arithmetics accordingly.

Answer to question 1

Answer to question 2

We are given the function g(x)=x2+4xg(x)=x2+4x.

Solving for the inverse, we have:

Solution:

Answer to question 3

Given:

4×2−49=04×2−49=0

This is a quadratic equation with:

a=4b=0c=−49a=4b=0c=−49

Since b=0b=0, we can apply the factorization formula:

4×2−49=(2x)2−72=(2x−7)(2x+7)=04×2−49=(2x)2−72=(2x−7)(2x+7)=0

There are two solutions:

2×1−7=02×2+7=02×1−7=02×2+7=0

The first equation produces a solution:

2×1=72×1=7

And the second one:

2×2=−72×2=−7

Thus, the two solutions are:

x=±72

Answer to question 4

Given:

f(x)=4×2+3x−7f(x)=4×2+3x−7

g(x)=x−1g(x)=x−1

a.) We solve for (fg)(x)(fg)(x) and simplify.

(fg)(x)=f(x)g(x)=4×2+3x−7x−1=(4x−7)(x−1)x−1Factor out the numerator=4x−7Cancel out the common term in the numerator and denominator (x-1)(fg)(x)=f(x)g(x)=4×2+3x−7x−1=(4x−7)(x−1)x−1Factor out the numerator=4x−7Cancel out the common term in the numerator and denominator (x-1)

Therefore, (fg)(x)(fg)(x) is equal to 4x−74x−7.

b.) We determine the domain of (fg)(x)(fg)(x) written in interval notation.

Since any real number can be the value of (x)(x) in the function (fg)(x)=4x−7(fg)(x)=4x−7 and will result in a defined value, hence the domain is all real numbers and written in interval notation as (−∞,∞)(−∞,∞).