Question

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

Answer 1

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

Explanation:

Decreasing function: A function is said to be decreasing function

When
`x_1<x_2`

Then,
`f(x_1)>f(x_2)`

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

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

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

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

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

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

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

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

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

Substitute x=-5

Then we get f(x)=0

Substitute x=-4

Then, we get

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

Substitute x=-3

Then, we get

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

When x increases then the value of function decrease.

Hence, the function is decreasing .

Option D is true.

Answer 2

We know

`y=\sqrt[3]{x}`

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

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

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

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

Only

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

is a decreasing function.

Option D) is the right answer.