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Range of f(x): All real numbers of y
Domain of g(x): All real numbers of x
Range g(x): y ≥ -7.5
f(x)&= g(x)
12(x-2)^3+1&= 2x^2-6x-3
From the diagram, we know these coordinates.
We have two solutions. A( 0,-3) &⇒ x= 0 B( 4,5) &⇒ x= 4 By substituting these x-values into the equation, we can show that they work. This means they make the left-hand side and right-hand side equal.
x= 0
Subtract term
Calculate power
1/b* a = a/b
Calculate quotient and product
Add terms
The first solution was correct. Let's also test the second solution.
x= 4
Subtract term
Calculate power
1/b* a = a/b
Calculate quotient and product
Add and subtract terms
The solutions are correct.
To determine if they do, we have to consider what happens when we go further left or further right in the coordinate plane.
To find the x-coordinate, we will rewrite the original equation so that it equals 0 and then factor it.
Examining the left-hand side, we notice that all of the terms contains at least one x This means we can factor the expression by pulling out this x. We will also factor out 0.5 which gives us nicer numbers inside of the parentheses. 0.5x(x^2-10x+24)=0 To find the third solution, we will use the Quadratic Formula on the trinomial inside of the parentheses.
Use the Quadratic Formula: a = 1, b= -10, c= 24
- (- a)=a
Calculate power and product
Subtract term
Calculate root
Calculate quotient
As we can see, we have one more solution at x=6.
This is only an approximation. To determine if we were correct, we should substitute it into f(x) and check if it makes the left-hand side and right-hand side equal.
x= 0.8, f(x)= 0
Subtract term
Calculate power
1/b* a = a/b
Calculate quotient
Add terms
Our estimation was not totally correct.
Looking at both f(x) and g(x), we notice that they go from left to right with no interruptions. This means there domain must be all real numbers of x. In other words, we can substitute any value of x and get an output. f(x):& All real numbers of x g(x):& All real numbers of x
Similar to the domain, f(x) does not appear to have any restrictions in the vertical direction. It comes from below and continues upward as we move to the right. Therefore its range is all real numbers of y.
.LHS /2.=.RHS /2.
Complete the square
Commutative Property of Addition
Split into factors
a^2-2ab+b^2=(a-b)^2
(a/b)^m=a^m/b^m
LHS-9/4=RHS-9/4
Calculate quotient
Subtract term
LHS * 2=RHS* 2
Add parentheses
Now we can identify the vertex. Function: & y=(x-1.5)^2+(-1.75) Vertex:& (1.5,-1.75) The vertex is at (1.5, -1.75) and since it is a minimum the range must be greater than or equal to -1.75. g(x): y≥-7.5