Pre Algebra Problems With Answers
Need help with some Pre Calculus and I think Algebra II problems?
Solve for x. Write the answer in interval notation:
-2 Less than or equal to 2x + 13/3 < 3
(the 3 divides the whole equation i.e. 2x+13 not just 13)
Write the domain in interval notation: Square root 2-9x
Assume x is less than 0, write the using radicals.
x ^ (3/5)
Given f(x) = 4x - 5 and g(x) = x2 + 3x. Find (g o f )(3).
Find the inverse function of the following:
f(x) = 6/8-x
Find the range in interval notation: g(x)= 5^x - 1
Solve: log(x+2) + log(x-1) = log(28)
Please answer, and if you can, please explain how you got the answer. This will REALLY help.
-2 Less than or equal to 2x + 13/3 < 3
-2 <= (2x+13)/3 < 3
(the 3 divides the whole equation i.e. 2x+13 not just 13)
-2 <= 2/3x+13/3 < 3 : subtract 13/3 from both sides
-2 - 13/3 <= (2/3)x < 3 - 13/3 : then multiply both sides by 3/2
(-19/3)*3/2 <= x < (-4/3) * 3/2
-19/2 <= x < -2: [-19/2 <= x < -2]
If complex numbers are not allowed then domain is where (2-9x) >=0
2-9x= 0, x = 2/9, so x <= 2/9: [- infinity, 2/9]
-x^3 * (1/5)sqrt(x) - 1/5 is the root of the radical
I used the properties of exponents: x^3.5 = x^3 * x^0.5
Given f(x) = 4x - 5 and g(x) = x2 + 3x. Find (g o f )
g(f(x)) = (4x-5)^2 + 3(4x - 5) =
16x^2 - 40x + 25 + 12x - 15 =
16x^2 - 28x +1 0
Find the inverse function of the following:
f(x) = 6/8-x
f(x) = y = 6/8 - x
inv(f) : x = 6/8 - y : simple case - just switch y and x
Find the range in interval notation: g(x)= 5^x - 1
[-1.0+, +infinity] the -1.0+ means that g(x) approaches -1.0 as x -> – infinity
Solve: log(x+2) + log(x-1) = log(28) : using properties of log and exponents:
1. log(b) + log(a) = log(ba)
log((x+2)(x-1))=log(28)
2.raise both sides to the power of e (or 10 whichever base your
log is:
e^log((x+2)(x-1))= e^log(28)
3. e^log(a) = a ->
(x+2)(x-1) = 28
x^2 + x – 30 = 0
Saxon Math Upper Grades
