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.

For each problem, find the: x and y intercepts, x-coordinates of the critical points, open intervals where the function is increasing and decreasing, x-coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. Using this information, sketch the graph of the function.

Problem 2 :

f(x) = -x³/3 + x²

Solution :

f(x) = -x³/3 + x²

Finding critical numbers :

f'(x) = -3x²/3 + 2x

f'(x) = -x² + 2x

f'(x) = 0

-x² + 2x = 0

-x(x - 2) = 0

x = 0 and x = 2

Critical numbers are 0 and 2.

Finding increasing and decreasing interval :

(-∞, 0) (0, 2) and (2, ∞)

Finding concavity on the intervals :

f'(x) = -x² + 2x

f''(x) = -2x + 2

f''(x) = 0

-2x + 2 = 0

-2x = -2

x = 1

(-∞, 1) and (1, ∞)

Point of inflection :

When x = 1 f''(x) = -2(1) + 2

+-1³/3 + 1

f''(x) = 0

-2x = -2

x = 1

Point of inflection at x = 1

Finding maximum/minimum :

Critical numbers should be applied in the second derivative

When x = 0

f''(0) = -2(0) + 2

= 0 (there is no max/min at x = 0)

When x = 2

f''(2) = -2(2) + 2

= -2 < 0 (maximum at x = 2)

When x = 2 in f(x)

f(2) = -2³/3 + 2²

= -8/3 + 4

= 4/3

So, maximum at (2, 4/3).

Problem 3 :

Sketch the graph of the function f(x) that satisfies f'(x) > 0 for x < -2 and x > 3 and f'(x) < 0 for -2 < x < 3

Solution :

• At x = -2, f'(x) changes from positive to negative there is local maximum.
• At x = 3, f'(x) changes from negative to positive there is local minimum.

Problem 4 :

Sketch the graph of f(x), assuming f(0) = 1 and given the graph of the derivative f'(x).

Solution :

x-intercepts are -2 and 0. Critical numbers are -2 and 0.

• At x = -2, slope changes from negative to positive. Then there is local minimum at x = -2.
• At x = 0, slope change from positive to negative. Then there is local maximum at x = 0.

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