Question

PLEASE HELP!!
Consider the functions f(x)=3x+30 and g(x)=x^2+20. At what positive integer value of x does the quadratic function, g(x) begin to exceed the linear function f(x)?

Answer 1

Answer:

$x\in(-\infty,-2)\cup(5,\infty)$

Step-by-step explanation:

The duadratic function $g(x)=x^2+20$ begin to exceed the linear function $f(x)=3x+30$ when $g(x)>f(x)$

Solve this inequality:

$x^2+20>3x+30\\ \\x^2-3x+20-30>0\\ \\x^2-3x-10>0\\ \\x^2-5x+2x-10>0\\ \\x(x-5)+2(x-5)>0\\ \\(x-5)(x+2)>0$

This inequality is equivalent to

$\left[\begin{array}{l}\left\{\begin{array}{l}x-5>0\\x+2>0\end{array}\right.\\ \\\left\{\begin{array}{l}x-5<0\\x+2<0\end{array}\right.\end{array}\right.\Rightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x>5\\x>-2\end{array}\right.\\ \\\left\{\begin{array}{l}x<5\\x<-2\end{array}\right.\end{array}\right.\Rightarrow \left[\begin{array}{l}x>5\\ \\x<-2\end{array}\right.$

Answer: $x\in(-\infty,-2)\cup(5,\infty)$

Answer 2

Answer:

x=5

Step-by-step explanation:

We are given that  

$f(x)=3x+30$

$g(x)=x^2+20$

We have to find the positive  integer value of x for which the quadratic function g(x) begin to exceed the linear function f(x).

$g(x) > f(x)$

$x^2+20 > 3x+30$

$x^2+20-3x-30 >0$

$x^2-3x-10 > 0$

$(x-5)(x+2) > 0$

$x-5 > 0$

$x > 5$

$x+2 > 0$

$x >-2$

Interval ($5,\infty)$

Therefore , g(x) exceed f(x) in the interval ($5,\infty)$.

g(x) begin to exceed the linear function at x=5