The function k(x) = (g x h)(x) is graphed below, where g is an exponential function and h is a linear function. If h(x) = x + 1, which option below give the formula for g? h(^x) = 2^x
h(^x) = –2^x
h(^x) = –2^–x
h(^x) = 2^–x

The function kx g x hx is graphed below where g is an exponential function and h is a linear function If hx x 1 which option below give the formula for g hx 2x class=

Respuesta :

h(x)=-2^-x, yeah this is definitely right


Answer:  The correct option is (D) [tex]g(x)=2^{-x}.[/tex]

Step-by-step explanation:  Given that the function k(x) = (g x h)(x) is graphed in the figure, where g is an exponential function and h is a linear function.

We are to find the formula for g, if h(x) = x + 1.

From the graph, we note the following two values :

[tex]k(0)=1,~~k(3)=0.5.[/tex]

Now, will check our options one by one.

Option (A) :

Here, [tex]g(x)=2^x.[/tex]

S0,

[tex]k(x)=(g\times h)(x)=g(x)\times h(x)=2^x(x+1).[/tex]

At x = 0 and 3, we get

[tex]k(0)=2^0(0+1)=1,\\\\k(3)=2^3(3+1)=32\neq 0.5.[/tex]

This option is not correct.

Option (B) :

Here, [tex]g(x)=-2^x.[/tex]

S0,

[tex]k(x)=(g\times h)(x)=g(x)\times h(x)=-2^x(x+1).[/tex]

At x = 0 and 3, we get

[tex]k(0)=-2^0(0+1)=-1\neq 1,\\\\k(3)=-2^3(3+1)=-32\neq 0.5.[/tex]

This option is not correct.

Option (C) :

Here, [tex]g(x)=-2^{-x}.[/tex]

So,

[tex]k(x)=(g\times h)(x)=g(x)\times h(x)=-2^{-x}(x+1).[/tex]

At x = 0 and 3, we get

[tex]k(0)=-2^{-0}(0+1)=-1\neq 1,\\\\k(3)=-2^{-3}(3+1)=-\dfrac{1}{8}(4)=-0.5\neq 0.5.[/tex]

This option is not correct.

Option (D) :

Here, [tex]g(x)=2^{-x}.[/tex]

S0,

[tex]k(x)=(g\times h)(x)=g(x)\times h(x)=2^{-x}(x+1).[/tex]

At x = 0 and 3, we get

[tex]k(0)=2^{-0}(0+1)=1,\\\\k(3)=2^{-3}(3+1)=\dfrac{1}{8}(4)=0.5.[/tex]

Therefore, [tex]g(x)=2^{-x}.[/tex]

This option is CORRECT.

Hence, (D) is the correct option.

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