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Find the range of the function f(x) = (x^2 + 14x + 9)/(x^2 + 2) as varies over all real numbers.

 Jan 21, 2021
 #1
avatar+11086 
+1

Find the range of the function f(x) = (x^2 + 14x + 9)/(x^2 + 2) as varies over all real numbers.

 

Hello Guest!

 

The function f (x) = (x ^ 2 + 14x + 9) / (x ^ 2 + 2) is a hyperbolic function with two extrema.

 

\( f(x) = (x^2 + 14x + 9)/(x^2 + 2) \)

\(\frac{df(x)}{dx}=\frac{u'v-uv'}{v^2}\\ \frac{df(x)}{dx}=\dfrac{(2x+14)(x^2+2)-(x^2+14x+9)(2x)}{(x^2+2)^2}=0\\ 2x^3+4x+14x^2+28-(2x^3+28x^2+18x)=0\\ 2x^3+4x+14x^2+28-2x^3-28x^2-18x=0\)

\(-14x^2-14x+28=0\\ x^2+x-2=0\)

\(x=-\dfrac{1}{2}\pm \sqrt{\frac{1}{4}+2}=-0.5\pm1.5\)

\(x \in \{-2,1\}\)

\(y=(x ^ 2 + 14x + 9) / (x ^ 2 + 2) \)

\(y_{min}=-2.5\\ y_{max}=8\)

 

The range of the function f(x) = (x^2 + 14x + 9)/(x^2 + 2)

as varies over all real numbers is

 

-2.5 < R < 8

laugh  !

 Jan 21, 2021
edited by asinus  Jan 21, 2021
edited by asinus  Jan 21, 2021
edited by asinus  Jan 21, 2021
edited by asinus  Jan 22, 2021
edited by asinus  Jan 22, 2021
 #2
avatar+267 
+2

Well, I'm late. But if other people is stuck on this problem and want a hint, here's one: 

 

the denominator (x^2+2 in this case) can't equal 0 because then it would be undefined. 

 Jan 21, 2021

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