Which statements are true for the functions g(x) = x2 and h(x) = -x ? Check all that apply.
For any value of x, g(x) will always be greater than h(x).
For any value of x, h(x) will always be greater than g(x).
g(x) > h(x) for x = -1.
g(x) For positive values of x, g(x) > h(x).
For negative values ofx, g(x) > h(x).

Question

10-02-2020
Mathematics

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Which statements are true for the functions g(x) = x2 and h(x) = -x ? Check all that apply.
For any value of x, g(x) will always be greater than h(x).
For any value of x, h(x) will always be greater than g(x).
g(x) > h(x) for x = -1.
g(x) For positive values of x, g(x) > h(x).
For negative values ofx, g(x) > h(x).

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SaniShahbaz SaniShahbaz · Answer 1 · 2020-02-18T03:41:38+01:00

Answer:

We conclude that the statement For positive values of x, g(x) > h(x) is TRUE.

Step-by-step explanation:

Considering the functions

[tex]g\left(x\right)\:=\:x^2[/tex]

[tex]h\left(x\right)\:=\:-x[/tex]

Checking the statement: For any value of x, g(x) will always be greater than h(x)

As

[tex]g\left(x\right)\:=\:x^2[/tex] represents the square of an independent variable x.

[tex]h\left(x\right)\:=\:-x[/tex] represents the negative of an independent variable x.

For example,

Putting x = 1 in [tex]g\left(x\right)\:=\:x^2[/tex] and [tex]h\left(x\right)\:=\:-x[/tex]

[tex]g\left(x\right)\:=\:x^2[/tex]

[tex]g\left(1\right)\:=\:\left(1\right)^2[/tex]

[tex]g\left(1\right)\:=\:1[/tex]

Also

[tex]h\left(x\right)\:=\:-x[/tex]

[tex]h\left(1\right)\:=\:-1[/tex]

But, if we put x = 0, both [tex]g\left(x\right)\:=\:x^2[/tex] and [tex]h\left(x\right)\:=\:-x[/tex] becomes zero.

And if we put x = -1, then again the values of both [tex]g\left(x\right)\:=\:x^2[/tex] and [tex]h\left(x\right)\:=\:-x[/tex] will be same which is 1.

Therefore, it is NOT TRUE that For any value of x, g(x) will always be greater than h(x).

Checking the statement: h(x) will always be greater than g(x)

This statement is false.

For example,

Putting x = 1 in [tex]g\left(x\right)\:=\:x^2[/tex] and [tex]h\left(x\right)\:=\:-x[/tex]

[tex]g\left(x\right)\:=\:x^2[/tex]

[tex]g\left(1\right)\:=\:\left(1\right)^2[/tex]

[tex]g\left(1\right)\:=\:1[/tex]

Also

[tex]h\left(x\right)\:=\:-x[/tex]

[tex]h\left(1\right)\:=\:-1[/tex]

So, putting x = 1 would bring [tex]g\left(x\right)\:=\:x^2[/tex] being greater than [tex]h\left(x\right)\:=\:-x[/tex].

Therefore, it is NOT TRUE that For any value of x, h(x) will always be greater than g(x).

Checking the statement: g(x) > h(x) for x = -1

Considering

[tex]g\left(x\right)\:=\:x^2[/tex]

[tex]h\left(x\right)\:=\:-x[/tex]

Putting x = -1 in [tex]g\left(x\right)\:=\:x^2[/tex] and [tex]h\left(x\right)\:=\:-x[/tex]

[tex]g\left(x\right)\:=\:x^2[/tex]

[tex]g\left(-1\right)\:=\:\left(-1\right)^2[/tex]

[tex]g\left(-1\right)\:=\:1[/tex]

Also

[tex]h\left(x\right)\:=\:-x[/tex]

[tex]h\left(-1\right)\:=\:-\left(-1\right)[/tex]

[tex]h\left(-1\right)\:=\:1[/tex]

So, when we put x = -1, [tex]g\left(x\right)=h\left(x\right)[/tex]

i.e. [tex]g\left(-1\right)=h\left(-1\right)[/tex]

Therefore, it is NOT TRUE that g(x) > h(x) for x = -1.

Checking the statement: For positive values of x, g(x) > h(x)

Since

[tex]g\left(x\right)\:=\:x^2[/tex] represents the square of an independent variable x.

[tex]h\left(x\right)\:=\:-x[/tex] represents the negative of an independent variable x.

So, for positive values of x, g(x) > h(x)

For example,

Putting x = 1 in [tex]g\left(x\right)\:=\:x^2[/tex] and [tex]h\left(x\right)\:=\:-x[/tex]

[tex]g\left(x\right)\:=\:x^2[/tex]

[tex]g\left(1\right)\:=\:\left(1\right)^2[/tex]

[tex]g\left(1\right)\:=\:1[/tex]

Also

[tex]h\left(x\right)\:=\:-x[/tex]

[tex]h\left(1\right)\:=\:-1[/tex]

Therefore, the statement For positive values of x, g(x) > h(x) is TRUE.

Checking the statement: For negative values of x, g(x) > h(x)

Considering

[tex]g\left(x\right)\:=\:x^2[/tex]

[tex]h\left(x\right)\:=\:-x[/tex]

Putting x = -1 in [tex]g\left(x\right)\:=\:x^2[/tex] and [tex]h\left(x\right)\:=\:-x[/tex]

[tex]g\left(x\right)\:=\:x^2[/tex]

[tex]g\left(-1\right)\:=\:\left(-1\right)^2[/tex]

[tex]g\left(-1\right)\:=\:1[/tex]

Also

[tex]h\left(x\right)\:=\:-x[/tex]

[tex]h\left(-1\right)\:=\:-\left(-1\right)[/tex]

[tex]h\left(-1\right)\:=\:1[/tex]

So, when we put x = -1, [tex]g\left(x\right)=h\left(x\right)[/tex]

Therefore, the statement For negative values of x, g(x) > h(x) is NOT TRUE.

Therefore, from the entire discussion above, we conclude that the statement For positive values of x, g(x) > h(x) is TRUE.

Keywords: function, composition function

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