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Which cube root function is always decreasing as x increases?


Af(x) = 3√x-8
Bf(x) = 3√x-5
Cf(x) = 3√-(5-x)
Df(x) = -3√x+5

Verified Answer :

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.

lublana

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.

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