Use this strategy to sketch the following graph, please. f(x) =( X + 1 ) \\ X^2

Question

Use this strategy to sketch the following graph, please.

f(x) =( X + 1 ) X^2 - 3x

Graph - sketching strategy to follow:

1.The domain of f

2. Whether f is even, odd or periodic( or none of these )

3. The x-intercepts and y-intercepts of f

4. The intervals which f is positive or negative (sign table for f(x) )

5. The intervals on which f is increasing or decreasing, the nature of any stationary points, and the value of f at each of these points. ( differentiate f(x) and sign table for f'(x) )

6. The asymptotic behaviour of f

Thanks for help !!

Explanation / Answer

Hi...

Given f(x)=(x+1)/(x^2-3x):

(1) The domain of a rational function is all real numbers except where the denominator is zero. Here x^2-3x=0 ==> x=0 or 3. So the domain is all real numbers except 0 and 3.

Since the numerator is nonzero at these points, there are vertical asymptotes at x=0 and x=3.

(2) The function is not even, odd, or periodic. Rational functions are not periodic. f(-x) is not equal to f(x) so the function is not even, and f(-x) is not equal to -f(x) so the function is not odd.

(3) There is no y-intercept as there is a vertical asymptote at x=0. There is an x-intercept at x=-1.

(4) We look at the intervals x<-1, -1<x<0,0<x<3, and x>3.

x<-1 ==> f(x)<0 (Numerator negative, denominator positive.)

-1<x<0 ==> f(x)>0 (Numerator positive, denominator positive.)

0<x<3 ==> f(x)<0 (Numerator positive, denominator negative.)

x>3 ==> f(x)>0 (Numerator positive, denominator positive.)

(5) Using the quotient rule f'(x)=-(x^2-2x+3)/(x^2-3x)^2

The derivative is zero at x=-3 and x=1. The derivative fails to exist at x=0. So the intervals of interest are x<-3, -3<x<0,0<x<1,x>1:

x<-3 ==> f'(x)<0 (Numerator negative) so the function decreases.

-3<x<0 ==> f'(x)>0 (Numerator positive) so the function increases.

0<x<1 ==> f'(x)>0 (Numerator positive) so the function increases.

x>1 ==> f'(x)<0 (Numerator negative) so the function decreases.

Note that the denominator is always positive.

(6) There are vertical asymptotes at x=0 and x=3.

For extreme values of x, the function acts like y=1/x so the horizontal asymptote is y=0. (The limit as x approaches +- infinity is 0.)

The graph:

(Note the x-intercept at x=-1. The function is negative for x<-1, and approaches y=0 from below.)

The stationary points are at x=-3 and x=1, because the derivative is zero at these points. (The derivative fails at x=0 and x=3, but the function is undefined there.) Each point is a local minimum, and x=-3 is a global minimum.

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