HOW TO SKETCH THE CURVE OF A FUNCTION

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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