determin wether true or false. (2 points) True False The functions f(x) = x – 5 and g(x) = -3x + 15 intersect at x = 5. The functions f (x) = 3 and g(x) = 11 – 2. intersect at x = 3. O The functions f (x) = x + 3 and g(x) = -x + 7 intersect at x = 2. The functions f (x) = {x – 3 and g(x) = -2x + 2 intersect at x = -2.



Answer :

To find the intersection point between f(x) and g(x) we will equate their right sides

[tex]\begin{gathered} f(x)=x-5 \\ g(x)=-3x+15 \end{gathered}[/tex]

Equate x - 5 by -3x + 15 to find x

[tex]x-5=-3x+15[/tex]

add 3x to both sides

[tex]\begin{gathered} x+3x-5=-3x+3x+15 \\ 4x-5=15 \end{gathered}[/tex]

Add 5 to both sides

[tex]\begin{gathered} 4x-5+5=15+5 \\ 4x=20 \end{gathered}[/tex]

Divide both sides by 4 to get x

[tex]\begin{gathered} \frac{4x}{4}=\frac{20}{4} \\ x=5 \end{gathered}[/tex]

Then the first one is TRUE

For the 2nd one

f(x) = 3, and g(x) = 11 - 2x

If x = 3, then substitute x by 3 in g(x)

[tex]\begin{gathered} g(3)=11-2(3) \\ g(3)=11-6 \\ g(3)=5 \end{gathered}[/tex]

Since f(3) = 3 because it is a constant function and g(x) = 5 at x = 3

That means they do not intersect at x = 3 because f(3), not equal g(3)

[tex]f(3)\ne g(3)[/tex]

Then the second one is FALSE

For the third one

f(x) = x + 3

at x = 2

[tex]\begin{gathered} f(2)=2+3 \\ f(2)=5 \end{gathered}[/tex]

g(x) = -x + 7

at x = 2

[tex]\begin{gathered} g(2)=-2+7 \\ g(2)=5 \end{gathered}[/tex]

Since f(2) = g(2), then

f(x) intersects g(x) at x = 2

The third one is TRUE

For the fourth one

[tex]f(x)=\frac{1}{2}x-3[/tex]

At x = -2

[tex]\begin{gathered} f(-2)=\frac{1}{2}(-2)-3 \\ f(-2)=-1-3 \\ f(-2)=-4 \end{gathered}[/tex]

g(x) = -2x + 2

At x = -2

[tex]\begin{gathered} g(-2)=-2(-2)+2 \\ g(-2)=4+2 \\ g(-2)=6 \end{gathered}[/tex]

Hence f(-2) do not equal g(-2), then

[tex]f(-2)\ne g(-2)[/tex]

f(x) does not intersect g(x) at x = -2

The fourth one is FALSE