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Which cube root function is always decreasing as x increases?

A) f(x) = 3√x-8
B) f(x) = 3√x-5
C) f(x) = 3√-(5-x)
D) f(x) = -3√x+5

Respuesta :

We know

[tex]y=\sqrt[3]{x}[/tex]

is an increasing function as when the value of x increases the value of y increases

And when the value of x decreases , the value of y also decreases.

Now if we have (x+a) or (x-a) instead of x, the function shall have a horizontal shift.

So it shall either move left or right but shall not flip.

So

[tex]y=\sqrt[3]{(x-8)}[/tex] and [tex]y=\sqrt[3]{(x-5)}[/tex]

are increasing functions.

Only when x becomes -x, that the function shall flip & shall become a decreasing function.

But then it must be - (x-a) or -(x+a) inside.

So

[tex]y=\sqrt[3]{-(5-x)}[/tex] is also increasing

Only

[tex]y=- \sqrt[3]{(x+5)}[/tex]

is a decreasing function.

Option D) is the right answer.

lublana

Answer:

D.[tex]f(x)=-\sqrt[3]{x+5}[/tex]

Step-by-step explanation:

Decreasing function: A function is said to be decreasing function

When [tex]x_1<x_2[/tex]

Then, [tex]f(x_1)>f(x_2)[/tex]

[tex]f(x)=\sqrt[3]{x}[/tex]

It is an increasing function because when x increases then the value of f(x) is also increases.

A.[tex]f(x)=\sqrt[3]{x-8}[/tex]

In given function only x- coordinate shift left side .Therefore, the cube root function remain increasing.

B.[tex]f(x)=\sqrt[3]{x-5}[/tex]

It is increasing function because when x increasing then the value of f(x) is also increase.

C.[tex]f(x)=\sqrt[3]{-(5-x)}[/tex]

It is also increasing function because when x increase then the value of function is also increases.

D.[tex]f(x)=-\sqrt[3]{x+5}[/tex]

Substitute x=-5

Then we get f(x)=0

Substitute x=-4

Then, we get

[tex]f(-4)=-\sqrt[3]{-4+5}=-1[/tex]

Substitute x=-3

Then, we get

[tex]f(-3)=-\sqrt[3]{-3+5}=-1.26[/tex]

When x increases then the value of function decrease.

Hence, the function is decreasing .

Option D is true.

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