Question

Which cube root function is always decreasing as x increases?

Answer 1

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.

Answer 2

Answer:

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

Step-by-step explanation:

Decreasing function: A function is said to be decreasing function

When $x_1$

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.