# Thread: Graphing using DERIVATIVES

1. ## Graphing using DERIVATIVES

I learned how to graph derivatives on Friday, but since I havent reviewed it i pretty much forgot EVERYTHING. Could you guys please post the steps to graphing a curve with the derivative correctly. I need to label the intercepts, relative extrema, points of inflection, and the asymptotes.

Also, could you solve an example problem lets say x^2 + 1 / x^2 - 9?

2. Originally Posted by PrecalcKid114
I learned how to graph derivatives on Friday, but since I havent reviewed it i pretty much forgot EVERYTHING. Could you guys please post the steps to graphing a curve with the derivative correctly. I need to label the intercepts, relative extrema, points of inflection, and the asymptotes.

Also, could you solve an example problem lets say x^2 + 1 / x^2 - 9?
Hello,

I'm not sure what you mean by "graphing a curve with the derivative correctly". So I'm going to show you what you can do as an examination with the given example:

$f(x)=\frac{x^2+1}{x^2-9}$

1. Domain: The denominator must be unequal zero. The denominator is zero at x = -3 or x = 3. Thus the domain is $D=\mathbb{R} \setminus \{-3, 3\}$

2. Asympotes: $\lim_{|x|\rightarrow \infty}f(x)=\lim_{|x|\rightarrow \infty}\left(\frac{x^2+1}{x^2-9}\right) = \lim_{|x|\rightarrow \infty}\left(\frac{x^2\left(1+\frac{1}{x^2}\right) }{x^2\left(1-\frac{9}{x^2}\right)}\right) = 1$
Thus you have 3 asymptotes: x = -3 or x = 3 or y = 1

3. Y-intercept: $f(0)=\frac{1}{-9}$

4. Range: $f(x)\leq\frac{1}{-9}~\text{ for all x in }~-3 < x < 3$ and $f(x) > 1~\text{ for }~|x| > 3$

That means the range is $R=\mathbb{R}\setminus \left\{y | -\frac{1}{9} < y < 1\right\}$

5. Zeros $f(x) \ne 0$. A fraction is zero if the numerator equals zero and the denominator unequals zero. The term $x^2+1$ will never become zero.

6. Relative extrema. Calculate first the first derivative of f:

$f'(x) = \frac{(x^2-9)\cdot 2x - (x^2+1) \cdot 2x}{(x^2-9)^2} =\frac{-20x}{(x^2-9)^2}$

$f''(x) = \frac{60(x^2+3)}{(x^2-9)^3}$ . Don't forget to cancel the factor (x²-9)

f'(x) = 0 ==> x = 0

$f''(0) = -\frac{20}{81}$ that means you have a relative maximum at x = 0

7. Point of inflection. Condition: $f''(x) = 0~~\wedge~~f'''(x) \ne 0$ . $f''(x) \ne 0$ (see my remark to #5) thus there is no point of inflection.

8. I've attached a sketch of the graph including the asymptotes.