Solving Equation II

f(x)=3x-4,g(x)=x+2

 

Setup the composite result function.  The fog notation is interpreted as f(g(x)).

f(g(x))

 

Evaluate f(g(x)) by substituting in the value of g into f.

f(x+2)=3(x+2)-4

 

Multiply 3 by each term inside the parentheses.

f(x+2)=3x+6-4

 

Subtract 4 from 6 to get 2.

f(x+2)=3x+2

 

(G*F)= f(x)=3x-4,g(x)=x+2

 

Multiply f by each term inside the parentheses.

f(x)=3x-4_g(x)=x+2_g(fx)

 

Multiply g by each term inside the parentheses.

f(x)=3x-4_g(x)=x+2_fgx

 

Compose the result function for fgx by replacing the function designators with the actual functions.

(3x-4)(x+2)x

 

Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group.

x(3x*x+3x*2-4*x-4*2)

 

Simplify the FOIL expression by multiplying and combining all like terms.

x(3x^(2)+2x-8)

 

Multiply x by each term inside the parentheses.

3x^(3)+2x^(2)-8x

 

Now (F*G)(2)=3*2+2 = 8

 

 

2.

F(x)=√x  and g(x)=x-1

 

f(x)=√(x)

 

The domain of an expression is all real numbers except for the regions where the expression is undefined.  This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0.  All of these are undefined and therefore are not part of the domain.

(x)<0

 

Solve the equation to find where the original expression is undefined.

x<0

 

The domain of the rational expression is all real numbers except where the expression is undefined.

[0, ,∞)

 

f(x)=x-1

 

The domain of the rational expression is all real numbers except where the expression is undefined.  In this case, there is no real number that makes the expression undefined.

All real numbers

 

 

3.

F(x)= f(x)=√(84-6x)

 

The domain of an expression is all real numbers except for the regions where the expression is undefined.  This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0.  All of these are undefined and therefore are not part of the domain.

(84-6x)<0

 

Solve the equation to find where the original expression is undefined.

x=14

 

The domain of the rational expression is all real numbers except where the expression is undefined.

(-∞,14) U (14,∞)

 

4.

F(x)=x^3+2

f(x)=x^(3)+2

 

To find the inverse of the function, interchange the variables and solve for f^(-1)(x).

x=f^(-1)(x)^(3)+2

 

Since f^(-1)(x) is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

f^(-1)(x)^(3)+2=x

 

Since 2 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 2 from both sides.

f^(-1)(x)^(3)=-2+x

 

Move all terms not containing f^(-1)(x) to the right-hand side of the equation.

f^(-1)(x)^(3)=x-2

 

Take the cube root of both sides of the equation to eliminate the exponent on the left-hand side.

 

 

f^(-1)(x)=√3:(x-2)

 

 

 

 

1.

Distance between points

(5,1),(8,5)

 

Use the distance formula to determine the distance between the two points.

distance=√((x2-x1)^(2)+(y2-y1)^(2))

 

Substitute the actual values of the points into the distance formula.

distance=√((8-(5))^(2)+(5-(1))^(2))

 

Multiply -1 by the 5 inside the parentheses.

distance=√((8-5)^(2)+(5-(1))^(2))

 

Subtract 5 from 8 to get 3.

distance=√((3)^(2)+(5-(1))^(2))

 

Expand the exponent (2) to the expression.

distance=√((3^(2))+(5-(1))^(2))

 

Squaring a number is the same as multiplying the number by itself (3*3).  In this case, 3 squared is 9.

distance=√((9)+(5-(1))^(2))

 

Multiply -1 by the 1 inside the parentheses.

distance=√((9)+(5-1)^(2))

 

Subtract 1 from 5 to get 4.

distance=√((9)+(4)^(2))

 

Expand the exponent (2) to the expression.

distance=√((9)+(4^(2)))

 

Squaring a number is the same as multiplying the number by itself (4*4).  In this case, 4 squared is 16.

distance=√((9)+(16))

 

Remove the parentheses that are not needed from the expression.

distance=√(9+16)

 

Add 16 to 9 to get 25.

distance=√(25)

 

Pull all perfect square roots out from under the radical.  In this case, remove the 5 because it is a perfect square.

distance=5 units

 

 

(10,4) and (2,6)

 

(10,4),(2,6)

 

Use the midpoint formula to find the midpoint of the line segment.

((x1+x2)/(2) , (y1+y2)/(2))

 

Substitute in the values for (x1,y1) and (x2,y2).

((10+2)/(2),(4+6)/(2))

 

Add 2 to 10 to get 12.

((12)/(2),(4+6)/(2))

 

Reduce the expression (12)/(2) by removing a factor of 2 from the numerator and denominator.

(6,(4+6)/(2))

 

Add 6 to 4 to get 10.

(6,(10)/(2))

 

Reduce the expression (10)/(2) by removing a factor of 2 from the numerator and denominator.

(6,5)

 

3.

Equation OF Circle

 

(x-(-3))^2+(y-5)^2=3^2

(x+9)^2+(y-5)^2=9

 

X^2+y^2 + 18x -10y + 97=0

 

 

4.

X^2+y^2+8x+4y+16=0

(X+4)^2 + Y^2 + 4Y + 4 – 4 =0

(X+4)^2+(Y+2)^2 = 2^2

 

 

Radius = 2 units

 

 

8.

 

f(x)=2x^(2)-7x-4

 

Replace f(x) with y to find the properties of the parabola.

y=2x^(2)-7x-4

 

To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of x.  In this problem, add (-(7)/(4))^(2) to both sides of the equation.

y=2(x^(2)-(7x)/(2)+(49)/(16))+2(-2)-(2)(0+(49)/(16))

 

Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression.

y=2(x^(2)-(7x)/(2)+(49)/(16))+2(-2)-(2)((49)/(16))

 

Factor the perfect trinomial square into (x-(7)/(4))^(2).

y=2((x-(7)/(4))^(2))+2(-2)-(2)((49)/(16))

 

Factor the perfect trinomial square into (x-(7)/(4))^(2).

y=2(x-(7)/(4))^(2)+2(-2)-(2)((49)/(16))

 

Multiply 2 by each term inside the parentheses.

y=2(x-(7)/(4))^(2)-4-(2)((49)/(16))

 

Multiply 2 by (49)/(16) to get (49)/(8).

y=2(x-(7)/(4))^(2)-4-((49)/(8))

 

Multiply -1 by the (49)/(8) inside the parentheses.

y=2(x-(7)/(4))^(2)-4-(49)/(8)

 

To add fractions, the denominators must be equal.  The denominators can be made equal by finding the least common denominator (LCD).  In this case, the LCD is 8.  Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.

y=2(x-(7)/(4))^(2)-4*(8)/(8)-(49)/(8)

 

Complete the multiplication to produce a denominator of 8 in each expression.

y=2(x-(7)/(4))^(2)-(32)/(8)-(49)/(8)

 

Combine the numerators of all fractions that have common denominators.

y=2(x-(7)/(4))^(2)+(-32-49)/(8)

 

Subtract 49 from -32 to get -81.

y=2(x-(7)/(4))^(2)+(-81)/(8)

 

Move the minus sign from the numerator to the front of the expression.

y=2(x-(7)/(4))^(2)-(81)/(8)

 

This is the form of a paraboloa.  Use this form to determine the values used to find vertex and x-y intercepts.

y=a(x-h)^(2)+k

 

Use the standard form to determine the vertex and x-y intercepts.

a=2_k=-(81)/(8)_h=(7)/(4)

 

The vertex of a parabola is (h,k).

Vertex: ((7)/(4),-(81)/(8))

 

Domainf(x)=2x^(2)-7x-4

 

The domain of the rational expression is all real numbers except where the expression is undefined.  In this case, there is no real number that makes the expression undefined.

All real numbers

f(x)=2x^(2)-7x-4

 

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

2x^(2)-7x-4=f(x)

 

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.

2x^(2)-7x-f(x)-4=0

 

Use the quadratic formula to find the solutions.  In this case, the values are a=2, b=-7, and c=-1f(x)-4.

x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

 

Use the standard form of the equation to find a, b, and c for this quadratic.

a=2, b=-7, and c=-1f(x)-4

 

Substitute in the values of a=2, b=-7, and c=-1f(x)-4.

x=(-(-7)\~((-7)^(2)-4(2)(-1f(x)-4)))/(2(2))

 

Multiply -1 by each term inside the parentheses.

x=(7\~((-7)^(2)-4(2)(-1f(x)-4)))/(2(2))

 

Simplify the section inside the radical.

x=(7\~(8f(x)+81))/(2(2))

 

Simplify the denominator of the quadratic formula.

x=(7\~(8f(x)+81))/(4)

 

First, solve the + portion of \.

x=(7+~(8f(x)+81))/(4)

 

Next, solve the – portion of \.

x=(7-~(8f(x)+81))/(4)

 

The final answer is the combination of both solutions.

x=(7+~(8f(x)+81))/(4),(7-~(8f(x)+81))/(4)

 

The domain of an expression is all real numbers except for the regions where the expression is undefined.  This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0.  All of these are undefined and therefore are not part of the domain.

(8f(x)+81)<0

 

Solve the equation to find where the original expression is undefined.

f(x)<-(81)/(8)

 

The domain of the rational expression is all real numbers except where the expression is undefined.

f(x)>=-(81)/(8)_[-(81)/(8),infinity

 

The domain of the inverse of f(x)=2x^(2)-7x-4 is equal to the range of f(f(x))=((7+~(8f(x)+81)))/(4).

Range: f(x)>=-(81)/(8)

=  [-(81)/(8),infinity]

 

 

 

2.

F(x) = 6- 4x+ x^2

f(x)=6-4x+x^(2)

 

Replace f(x) with y to find the properties of the parabola.

y=6-4x+x^(2)

 

Nothing

y=-4(-(x^(2))/(4)+x)

 

Nothing

y=4(-(x^(2))/(4)*-4+x*-4)-(-4)(0*-4)

 

 

Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression.

y=-4(x^(2)-4x+4)-(-4)(4)

 

Factor the perfect trinomial square into (x-2)^(2).

y=-4((x-2)^(2))-(-4)(4)

Factor the perfect trinomial square into (x-2)^(2).

y=-4(x-2)^(2)-(-4)(4)

 

Multiply -4 by 4 to get -16.

y=-4(x-2)^(2)-(-16)

 

Multiply -1 by each term inside the parentheses.

y=-4(x-2)^(2)+16

 

This is the form of a paraboloa.  Use this form to determine the values used to find vertex and x-y intercepts.

y=a(x-h)^(2)+k

 

Use the standard form to determine the vertex and x-y intercepts.

a=-4_k=16_h=2

 

The vertex of a parabola is (h,k).

Vertex: (2,16)

 

 

 

 

Graph

f(x)=6-4x+x^(2)

 

The domain of the rational expression is all real numbers except where the expression is undefined.  In this case, there is no real number that makes the expression undefined.

All real numbers

f(x)=6-4x+x^(2)

 

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

6-4x+x^(2)=f(x)

 

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.

x^(2)-4x-f(x)+6=0

 

Use the quadratic formula to find the solutions.  In this case, the values are a=1, b=-4, and c=-1f(x)+6.

x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

 

Use the standard form of the equation to find a, b, and c for this quadratic.

a=1, b=-4, and c=-1f(x)+6

 

Substitute in the values of a=1, b=-4, and c=-1f(x)+6.

x=(-(-4)\~((-4)^(2)-4(1)(-1f(x)+6)))/(2(1))

 

Multiply -1 by each term inside the parentheses.

x=(4\~((-4)^(2)-4(1)(-1f(x)+6)))/(2(1))

 

Simplify the section inside the radical.

x=(4\2~(-1(-f(x)+2)))/(2(1))

 

Simplify the denominator of the quadratic formula.

x=(4\2~(-1(-f(x)+2)))/(2)

 

First, solve the + portion of \.

x=(4+2~(-1(-f(x)+2)))/(2)

 

Simplify the expression to solve for the + portion of the \.

x=2+~(-1(-f(x)+2))

 

Next, solve the – portion of \.

x=(4-2~(-1(-f(x)+2)))/(2)

 

Simplify the expression to solve for the – portion of the \.

x=2-~(-1(-f(x)+2))

 

The final answer is the combination of both solutions.

x=2+~(-1(-f(x)+2)),2-~(-1(-f(x)+2))

 

The domain of an expression is all real numbers except for the regions where the expression is undefined.  This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0.  All of these are undefined and therefore are not part of the domain.

(-1(-f(x)+2))<0

 

Solve the equation to find where the original expression is undefined.

f(x)<2

 

The domain of the rational expression is all real numbers except where the expression is undefined.

f(x)>=2_[2,infinity

 

The domain of the inverse of f(x)=6-4x+x^(2) is equal to the range of f(f(x))=2+~(-1(-f(x)+2)).

Range: f(x)>=2_[2,infinity)

 

3.

 

f(x)=-3*(x+(1)/(2))(x-4)^(3)

 

To find the roots)/(zeros of the function, set the function equal to 0 and solve.

0=-3*(x+(1)/(2))(x-4)^(3)

 

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

-3*(x+(1)/(2))(x-4)^(3)=0

 

Multiply -3 by (x+(1)/(2))(x-4)^(3) to get -3(x+(1)/(2))(x-4)^(3).

-3(x+(1)/(2))(x-4)^(3)=0

 

If any individual factor on the left-hand side of the equation is equal to 0, the entire expression will be equal to 0.

(x+(1)/(2))=0_(x-4)^(3)=0

 

Set the first factor equal to 0 and solve.

(x+(1)/(2))=0

 

Remove the parentheses around the expression x+(1)/(2).

x+(1)/(2)=0

 

Since (1)/(2) does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting (1)/(2) from both sides.

x=-(1)/(2)

 

Set the next factor equal to 0 and solve.

(x-4)^(3)=0

 

Set each of the factors of the left-hand side of the equation equal to 0.

x-4=0

 

Since -4 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 4 to both sides.

x=4

 

The final solution is all the values that make -3(x+(1)/(2))(x-4)^(3)=0 true.  The multiplicity of a root is the number of times the root appears.  For example, a factor of (x-4)^(3)=0 has multiplicity of 3.

x=-(1)/(2),4 with multiplicity: 3

 

 

the graph  crosses the x axis

 

 

4.

Leading co-efficient test

F(x)=-2*x^4+2*x^3

When the leading coefficient is negative and the degree is even, it falls to the left and falls to the right

 

f(F,x)=-2*x^(4)+2*x^(3)

 

To find the x-intercept, substitute in 0 for y and solve for x.

(0)=-2*x^(4)+2*x^(3)

 

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

-2*x^(4)+2*x^(3)=(0)

 

Multiply -2 by x^(4) to get -2x^(4).

-2x^(4)+2*x^(3)=(0)

 

Multiply 2 by x^(3) to get 2x^(3).

-2x^(4)+2x^(3)=(0)

 

Remove the parentheses around the expression 0.

-2x^(4)+2x^(3)=0

 

Multiply each term in the equation by -1.

2x^(4)-2x^(3)=0

 

Factor out the GCF of 2x^(3) from each term in the polynomial.

2x^(3)(x)+2x^(3)(-1)=0

 

Factor out the GCF of 2x^(3) from 2x^(4)-2x^(3).

2x^(3)(x-1)=0

 

Set the single term factor on the left-hand side of the equation equal to 0.

2x^(3)=0

 

Divide each term in the equation by 2.

(2x^(3))/(2)=(0)/(2)

 

Simplify the left-hand side of the equation by canceling the common factors.

x^(3)=(0)/(2)

 

0 divided by any number or variable is 0.

x^(3)=0

 

Take the cube root of both sides of the equation to eliminate the exponent on the left-hand side.

x=cubic root(0)

 

Pull all perfect cube roots out from under the radical.  In this case, remove the 0 because it is a perfect cube.

x=0

 

Set each of the factors of the left-hand side of the equation equal to 0.

x-1=0

 

Since -1 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 1 to both sides.

x=1

 

The complete solution is the set of the individual solutions.

x=0,1

 

To find the y-intercept, substitute in 0 for x and solve for y.

y=-2*(0)^(4)+2*(0)^(3)

 

Expand the exponent (4) to the expression.

y=-2*(0^(4))+2*(0)^(3)

 

Raising a number to the 4th power is the same as multiplying the number by itself 4 times.  In this case, 0 raised to the 4th power is 0.

y=-2*(0)+2*(0)^(3)

 

Expand the exponent (3) to the expression.

y=-2*(0)+2*(0^(3))

 

Cubing a number is the same as multiplying the number by itself 3 times (0*0*0).  In this case, 0 cubed is 0.

y=-2*(0)+2*(0)

 

Multiply all the factors separated by a * in -2*(0)+2*(0).

y=-2(0)+2(0)

 

Multiply -2 by each term inside the parentheses.

y=0+2(0)

 

Multiply 2 by each term inside the parentheses.

y=0+0

 

Solve the equation.

y=0

 

These are the x and y intercepts of the equation y=-2*x^(4)+2*x^(3).

x=0,1, y=0

 

 

 

 

We can see the graph has no symmetry