Analyzing Graph of Function

Analyzing Graph of Function : To analyze the graph of a given function we use the derivatives. We use various technique to analyze the behavior of the graph. Let us summarize the result of analyzing graph using the following table.
• Domain and range
• x and y intercepts
• Symmetry
• Continuity
• Vertical asymptotes
• Horizontal asymptotes
• (critical points)Differentiability
• Relative extrema
• Concavity
• Points of inflection
• Infinite limits at infinity
After following the above steps it will be easy to sketch any given function without using a graphing calculator.

Examples on Analyzing Graph of Function

Example 1: f(x) = $\frac{x^{2}-4}{x^{2} -25}$

Solution :
Step I : Find the domain
As this is a rational function so to find the domain, set the denominator equal to zero
$x^{2}$ - 25 = 0
$x^{2}$ = 25
x = $\pm$ 5
So $D_{f} = { x| x \in R, x \neq \pm 5}$
OR interval notation is ($-\infty$, 5) U (-5,5) U (5,$\infty$)
Range is the output of the function, $R_{f} = (-\infty, 4/25] U (1,\infty)$

Step II : Check the given function is odd, even or neither

f(x) = $\frac{x^{2}-4}{x^{2} -25}$

f(-x) = $\frac{(-x)^{2}-4}{(-x)^{2} -25}$

= $\frac{x^{2}-4}{x^{2} -25}$

f(-x) = f(x) so the function is even.

Step III : Find x and y intercepts
To find x-intercept , plug in y=0 in the given function
f(x) = $\frac{x^{2}-4}{x^{2} -25}$

0 = $\frac{x^{2}-4}{x^{2} -25}$

$x^{2}$-4 = 0
$x^{2}$ = 4
x = $\pm 2$
X-intercept : (-2,0) and (2,0)
To find the y-intercept, plug in x = 0 in the given function
f(x) = $\frac{x^{2}-4}{x^{2} -25}$

f(0) = $\frac{0^{2}-4}{0^{2} -25}$

y = $\frac{-4}{-25}$

y = 4/25
Y-intercept : (0,4/25)

Step IV : Find vertical asymptotes
For vertical asymptotes, set the denominator equal to zero
$x^{2}$ - 25 = 0
$x^{2}$ = 25
x = $\pm$ 5
Vertical asymptote : x = 5 and x = -5
Step V : Find horizontal asymptotes As the degree of the numerator is equal to the degree of denominator then the horizontal asymptote is

y = $\frac{Leading coefficient of numerator}{Leading coefficient of denominator} $

so Horizontal asymptote : y = 1
Step IV : Find the derivatives for( differentiability, Relative extrema, Concavity, Points of inflection)
f(x) = $\frac{x^{2}-4}{x^{2} -25}$

=$\frac{(x^{2} -25)\frac{ \text{d}}{\text{d}x}(x^{2} -4)-(x^{2} -4)\frac{\text{d}}{\text{d}x}(x^{2}-25)}{(x^{2}-25)^{2}}$

=$=\frac{(x^{2} -25)(2x)-(x^{2} -4)(2x)}{(x^{2}-25)^{2}}$

f '(x) = $\frac{-42x}{(x^{2}- 25)^{2}}$

f "(x) = $ \frac{42(3x^{2}+25) }{(x^{2}- 25)^{3}}$

For critical points f '(x) = 0
$\frac{-42x}{(x^{2}- 25)^{2}}$ = 0

-42x = 0
x= 0
f '(x) is undefined at x = $\pm$ 5
Interval for critical points : ($\infty$, -5) (-5,0)(0,5) (5,$\infty$ )
Set the table :

	(-∞,-5)   ,-5)	x=-5	(-5,0)	x=0	(0,5)	x=5	(5,∞)
f(x)		undf		4/25		undf
f '(x)	+	undf	+	0	-	undf	-
f "(x)	+	undf	-	-	-	undf	+
Conclusion	increasing		Increasing		decreasing		decreasing
	concave up		concave down		concave down		concave up
				Rel.max

Point of inflection f "(x) $\neq$ 0

12th grade math

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