HOW TO SKETCH THE CURVE OF A FUNCTION

When we are sketching the graph of functions either by hand or through any graphing software we cannot show the entire graph. Only a part of the graph can be sketched.

A crucial question is which part of the curve we need to show and how to decide that part. To decide on this we use the derivatives of functions. We enlist few guidelines for determining a good viewing rectangle for the graph of a function. They are :

(i) The domain and the range of the function.

(ii) The intercepts of the cure (if any).

(iii) Critical points of the function.

(iv) Local extrema of the function.

(v) Intervals of concavity.

(vi) Points of inflections (if any).

(vii) Asymptotes of the curve (if exists)

Sketch the graphs of the following functions.

Problem 1 :

y  =  (-1/3)(x3-3x+2)

Solution :

Domain and range :

Since it is a polynomial function, all real values are domain and range.

Intercepts :

To find x intercept, put y  =  0

x3-3x+2  =  0

Factors are  =  (x-1) (x2+x-2) 

=  (x-1) (x+ 2)(x-1)

Equating each to zero, we get x  =  1, -2 and 1.

So, 1 and -2 are x-intercepts.

Y-intercept :

x  =  0

y  =  -2/3

Critical points :

f(x)  =  (-1/3)(x3-3x+2)

f'(x)  =  (-1/3) (3x2-3)

f'(x)  =  -1(x2-1)

f'(x)  =  -1(x+1)(x-1)

f'(x)  =  0

-1(x+1)(x-1)  =  0

x  =  -1, 1

Plotting these points on the number line and dividing into intervals, we get (-∞, -1), (-1, 1) and (1, ∞).

Local extrema :

f'(x)  =  -1(x2-1)

f''(x)  =  -2x

f''(-1)  =  2 > 0 (Minimum)

f''(1)  =  -2 < 0 (Maximum)

Maximum value at x  =  1 :

y  =  (-1/3)(13-3(1)x+2)

y  =  (-1/3)(1-3+2)

y  =  0

Minimum value at x  =  -1 :

y  =  (-1/3)(-13-3(-1)+2)

y  =  (-1/3)(-1+3+2)

y  =  4/3

Intervals of concavity :

f''(x)  =  -2x

f''(x)  =  0

2x  =  0

x  =  0

Dividing this into intervals, we get

(-∞, 0) and (0, ∞)

Point of inflection :

At x  =  0

y  =  (-1/3)(03-3(0)+2)

y  =  -2/3

So, point of inflection is (0, -2/3).

Asymptotes :

It will have only slant asymptote.

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