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)$

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

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

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

$x^2+5>25x^2$

$1+\frac{5}{x^2}>25$

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

Consider the function

$k(x)=(\frac{5}{2})^x$

Comparing the function with f(x), we get

$(\frac{5}{2})^x>25x^2$

Taking log on both sides, we get

$xIn\frac{5}{2}^{}>2In5x$

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

$\frac{1^k}{5k}^{}>2$

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).