Question

Select all the functions whose output values will eventually overtake the output values of function f defined by f(x) = 25x^2

Answer 1

Given function is

`f(x)=25x^2`

Eventually overtake means the function gets the value of f(x) at some value of x.

Consider the first option

`g(x)=5(2)^x`

Comparing the functions

`5(2)^x>25x^2`

`(2)^x>5x^2`
`xIn\text{ 2>2In (}\sqrt[]{5}x)`

`\frac{x}{\text{In (}\sqrt[]{5}x)}>(2)/(In2)`

`\frac{x}{\text{In (}\sqrt[]{5}x)}>1`

It is true, hence g(x) eventually overtakes f(x).

Consider the function

`h(x)=5^x`

It is increasing function, it also eventually overtakes the function f(x).

Consider the function

`j(x)=x^2+5`

Comparing the function with f(x), we get

`x^2+5>25x^2`

`1+(5)/(x^2)>25`

This is not true, hence j(x) is not overtaking the function f9x).

Consider the function

`k(x)=((5)/(2))^x`

Comparing the function with f(x), we get

`((5)/(2))^x>25x^2`

Taking log on both sides, we get

`xIn(5)/(2)^{}>2In5x`

`(x)/(In5x)^{}>(2)/(In((5)/(2)))`
`\text{Let x=}(1^k)/(5),\text{ we get}`

`(1^k)/(5k)^{}>2`

It is not true for any k value, so the function k(x) is not overtaking the function f(x).

Consider the function

`m(x)=5+2^x`

Comparing the function with f(x), we get

`5+2^x>25x^2`

`2^x>(5x)^2-5`

It is true for some value of x.

Hence the function m(x) overtakes the function f(x).

Consider the function

`n(x)=2x^2+5`

Comparing the function with f(x), we get

`2x^2+5>25x^2`

`5>23x^2`

It is not true for any value of x.

Hecne the function n(x) not overtake the function f(x).

Hence the following functions are overtaken f(x).

g(x), h(x), m(x).