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algol13
1 month ago
15

Which statements are true for the functions g(x) = x2 and h(x) = –x2 ? Check all that apply.

Mathematics
2 answers:
Leona [12.6K]1 month ago
6 0
If x equals zero, both functions have the same value.

Therefore, the first and second choices are incorrect.

Evaluating at x = -1:
g(-1) equals 1,
h(-1) equals -1,
so the third option is correct.

The fourth option is false.

Except at zero, for all other values, g(x) is greater than h(x).


Hence, the valid statements are:

g(x) is greater than h(x) when x = -1.
For positive x values, g(x) exceeds h(x).
For negative x values, g(x) exceeds h(x).
Svet_ta [12.7K]1 month ago
5 0

Answer: g(x) > h(x) when x = -1.

For x greater than zero, g(x) is greater than h(x).

For x less than zero, g(x) is greater than h(x).

Step-by-step explanation:

Given the functions: g(x)=x^2 and h(x)=-x^2

At x = 0, g(0)=0^2=0 and h(0)=-0^2=0

. Therefore, g(0) equals h(0).

Hence, statements claiming that for all x, g(x) is always greater than h(x), or vice versa, are incorrect.

At x = -1, g(-1)=(-1)^2=1 and h(-1)=-(-1)^2=-1

So, g(x) > h(x) for x = -1............................(1)

At x = 3, g(3)=(3)^2=9 and h(3)=-(3)^2=--9

Therefore, g(x) > h(x) at x = 3........................(2)

Which disproves the statement that g(x) < h(x) for x = 3.

From points (1) and (2):

For any positive x, g(x) > h(x).

For any negative x, g(x) > h(x).

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Two functions are shown in the table below. Function 1 2 3 4 5 6 f(x) = −x2 + 4x + 12 g(x) = −x + 6 Complete the table on your o
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For \fbox{\begin \\\math{x}=6\\\end{minispace}} the function f(x)=-x^{2} +4x+12 and g(x)=-x+6 both yield the same result.

Detailed breakdown:  

The functions involved are

f(x)=-x^{2}+4x+12

g(x)=-x+6

Step 1:  

Insert x=1 in f(x)=-x^{2} +4x+12 to find the value of f(1).

f(1)=-1^{2} +4(1)+12\\f(1)=-1+4+12\\f(1)=15

Insert x=1 in g(x)=-x+6 to find the value of g(1).

g(1)=-1+6\\g(1)=5

Step 2:

Insert x=2 in f(x)=-x^{2} +4x+12 to obtain the value of f(2).

f(2)=-2^{2} +4(2)+12\\f(2)=-4+8+12\\f(2)=16

Substitute x=2 into g(x)=-x+6 to find the value of g(2).

g(2)=-2+6\\g(2)=4

Step 3:

Replace x=3 in f(x)=-x^{2} +4x+12 to find the value of f(3).

f(3)=-3^{2} +4(3)+12\\f(3)=-9+12+12\\f(3)=15

Also, replace x=3 in g(x)=-x+6 to find the value of g(3).

g(3)=-3+6\\g(3)=3

Step 4:

Insert x=4 in f(x)=-x^{2} +4x+12 to find the value of f(4).

f(4)=-4^{2} +4(4)+12\\f(4)=-16+16+12\\f(4)=12

Also, replace x=4 in g(x)=-x+6 to obtain the value of g(4).

g(4)=-4+6\\g(4)=2

Step 5:

Insert x=5 in f(x)=-x^{2} +4x+12 to obtain the value of f(5).

f(5)=-5^{2} +4(5)+12\\f(5)=-25+20+12\\f(5)=7

Replace x=5 in g(x)=-x+6 to find the value of g(5).

g(5)=-5+6\\g(5)=1

Step 6:

Insert x=6 into f(x)=-x^{2} +4x+12 to find the value of f(6).

f(6)=-6^{2} +4(6)+12\\f(6)=-36+24+12\\f(6)=0

Also, substitute x=6 in g(x)=-x+6 to obtain the value of g(6).

g(6)=-6+6\\g(6)=0

Step 7:

According to the provided condition f(x)=g(x).

(a). Insert f(x)=-x^{2} +4x+12 and g(x)=-x+6 into the previously mentioned equation.

-x^{2} +4x+12=-x+6

(b). Multiply through by -1 on both sides.

x^{2} -4x-12=x-6

(c). Move the term x-6 to the left side of the equation.

x^{2} -4x-12-x+6=0\\x^{2} -5x-6=0

(d). Divide the middle term so that its sum equals 5 and the product equals 6.

x^{2} -(6-1)x-6=0\\x^{2} -6x+x-6=0\\x(x-6)+1(x-6)=0\\(x+1)(x-6)=0\\x=-1,6

From the analysis above, it is noted that for x=6 both functions f(x) and g(x) yield the same outcome.

Using a direct approach:

f(x)=g(x)\\\Leftrightarrow-x^{2} +4x+12=-x+6\\\Leftrightarrow-x^{2} +4x+12+x-6=0\\\Leftrightarrow-x^{2} +5x+6=0\\\Leftrightarrow-x^{2} +6x-x+6=0\\\Leftrightarrow x^{2} -6x+x-6=0\\\Leftrightarrow x(x-6)+1(x-6)=0\\\Leftrightarrow(x+1)(x-6)=0\\\Leftrightarrow x=6,-1

The table representing function f(x)=-x^{2} +4x+12 and g(x)=-x+6 is included below.

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