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The question is to find where the graph is increasing or decreasing.

The original function is $f(x)={(x^2+2x-48)}/{x^2}$

I know I need to find the prime of this function and I think it is this after using the quotient rule:


Finally, in order to draw a graph I need to find the points on the x axis that are either undefined and/or the slope equals 0.

Immediately I know that 0 is an undefined point because it makes the prime graph undefined.

It's when I take the numerator and set it equal to 0 that trips me up.




I'm guessing I did something wrong finding this prime because the answers are all including the number 48. I'm stumped because I've been at this question for about 1 hour.


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Please edit the title so that it becomes useful. – Mariano Suárez-Alvarez Jun 5 '12 at 6:06
How's that? :) . – ninja08 Jun 5 '12 at 6:08
If $T$ is the top and $B$ the bottom, then the derivative of $\frac{T}{B}$ is $\frac{T'B-TB'}{B^2}$. Simplify the top of this carefully. – André Nicolas Jun 5 '12 at 6:12
Thats exactly how I worked it out originally. I just missed something along the way. I don't know what I messed up though. The calc isn't hard it's the algebra that trips me up. – ninja08 Jun 5 '12 at 6:17
Feel free to roll it back if you don't like the title I've given. – Gerry Myerson Jun 5 '12 at 6:39
up vote 1 down vote accepted

You computation of derivative is incorrect. It is easier to separate out each term and compute the derivative. $$f(x) = \dfrac{x^2 + 2x - 48}{x^2} = 1 + \dfrac2{x} - \dfrac{48}{x^2}$$ Hence, $$f'(x) = 0 - \dfrac2{x^2} + \dfrac{2 \times 48}{x^3} = \dfrac{96-2x}{x^3}$$ Now setting, $f'(x) = 0$ gives us $96-2x = 0 \implies x = 48$.


As FrankScience rightly points out in his comment, you do not really need to differentiate to figure out the behavior of this function. We have that \begin{align} f(x) & = \dfrac{x^2 + 2x - 48}{x^2}\\ & = -48 \left( \left(\dfrac1x \right)^2 - \dfrac1{24} \dfrac1x\right) + 1\\ & = -48 \left( \left(\dfrac1x \right)^2 - 2 \dfrac1x\dfrac1{48} + \left(\dfrac1{48} \right)^2 - \left(\dfrac1{48} \right)^2\right) + 1\\ & = -48 \left( \dfrac1x - \dfrac1{48}\right)^2 + \dfrac1{48} + 1\\ & = \dfrac{49}{48} -48 \left( \dfrac1x - \dfrac1{48}\right)^2 \end{align}

Hence, note that the functions hits the maximum at $x=48$, since for all $x \neq 48$, you always subtract a positive quantity from $\dfrac{49}{48}$.

Analyze what happens when $x \to \infty$, $x \to - \infty$, $x \to 0^{\pm}$, $x \to 48^{\pm}$ to draw conclusions.

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Thanks! I'm working it out now. – ninja08 Jun 5 '12 at 6:10
Quadratic function can be solved without derivative or differentiating. – Frank Science Jun 5 '12 at 6:44
@FrankScience True. I was just pointing out where the OP had made a mistake. – user17762 Jun 5 '12 at 6:46
The way you split up the function helps a lot. In the second explanation I have no idea what you did. I'd look into it deeper if I had more time. Thank you! – ninja08 Jun 5 '12 at 14:50

When you say "find the prime of this function" you presumably mean "find the derivative of this function"

The quotient rule $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$ should have given you

$$ f'(x) = \frac{(2x+2)x^2 - (x^2+2x-48)2x}{x^4} = \frac{96-2x}{x^3}$$ as Marvis found by a different route.

$f'(x)$ is clearly negative when $x$ is negative (the numerator is positive and the denominator negative) and when $x \gt 48$ (the numerator is negative and the denominator positive).

$f'(x)$ is clearly positive when $0 \lt x \lt 48$ (the numerator is positive and the denominator positive).

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You may use this form and solve it elementarily:

$$f(x) = \dfrac{x^2 + 2x - 48}{x^2}=\dfrac{((x+1)-7)((x+1)+7)}{x^2}=\dfrac{((x+1)^2-49)}{x^2}=\left(\dfrac{x+1}{x}\right)^2 - \frac{49}{x^2}$$

After denoting $\dfrac{x+1}{x}=y$, we get $-48 y^2+98 y-49$ that has as a maximum the value $\frac{49}{48}$ at $y=\frac{49}{48}$, and hence $f$ reaches its maximum at $x=48$. According to our function interpretation we see that$f$ is strictly decreasing on $(-\infty,0)$ and $(48,+\infty)$ and strictly increasing on $(0,48)$.

The proof is complete.

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Ahh. It must have been the factoring that messed me up. I didn't factor the numerator. I haven't factored that kind of poly in such a long time. Thanks the the help! – ninja08 Jun 5 '12 at 14:51

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