Derivatives and the shapes of graphs

Curve SketchingBy

Sharda Haryani

Derivatives and the shapes of graphs

Increasing / Decreasing Test:(a) If f ′ (x) > 0 on an interval, then f is increasing

on that interval.(b) If f ′ (x) < 0 on an interval, then f is decreasing

on that interval.

Example: Find where the function f (x) = x3 – 1.5x2 – 6x + 5 is increasing and where it is decreasing.

Solution: f ′ (x) = 3x2 – 3x – 6 = 3(x + 1)(x - 2) f ′ (x) > 0 for x < -1 and x > 2 ; thus the function is increasing on (-, -1) and (2, ) . f ′ (x) < 0 for -1 < x < 2 ;

thus the function is decreasing on (-1, 2) .

The First Derivative Test: Suppose that c is a critical number of a continuous function f.

(a) If f ′ is changing from positive to negative at c, then f has a local maximum at c.

(b) If f ′ is changing from negative to positive at c, then f has a local minimum at c.

(c) If f ′ does not change sign at c, then f has no local maximum or minimum at c.

Example(cont.): Find the local minimum and maximum values of the function f (x) = x3 – 1.5x2 – 6x

+ 5.

Solution: f ′ (x) = 3x2 – 3x – 6 = 3(x + 1)(x - 2)

f ′ is changing from positive to negative at -1 ; so f (-1) = 8.5 is a local maximum value ;

f ′ is changing from negative to positive at 2 ; so f (2) = -5 is a local minimum value.

Concave upward and downwardDefinition:

(a) If the graph of f lies above all of its tangents on an interval, then f is called concave upward on that interval.

(b) If the graph of f lies below all of its tangents on an interval, then f is called concave downward on that interval.

Concave upward Concave downward

Inflection PointsDefinition:

A point P on a curve y = f(x) is called an inflection point if f is continuous there and the curve changes

• from concave upward to concave downward or • from concave downward to concave upward at P.

Inflection points

Concavity test:(a) If f ′ ′ (x) > 0 for all x of an interval, then the graph of

f is concave upward on the interval.(b) If f ′ ′ (x) < 0 for all x of an interval, then the graph of

f is concave downward on the interval.

Example(cont.): Find the intervals of concavity of the function f (x) = x3 – 1.5x2 – 6x + 5.

Solution: f ′ (x) = 3x2 – 3x – 6 f ′ ′ (x) = 6x - 3 f ′ ′ (x) > 0 for x > 0.5 , thus it is concave upward on (0.5, ) . f ′ ′ (x) < 0 for x < 0.5 , thus it is concave downward on (-, 0.5) .Thus, the graph has an inflection point at x = 0.5 .

What does f ′ ′ say about f ?

The second derivative test: Suppose f is continuous near c.

(a) If f ′ (c) = 0 and f ′ ′ (c) > 0 then f has a local minimum at c.

(b) If f ′ (c) = 0 and f ′ ′ (c) < 0 then f has a local maximum at c.

Example(cont.): Find the local extrema of the function f (x) = x3 – 1.5x2 – 6x + 5.

Solution: f ′ (x) = 3x2 – 3x – 6 = 3(x + 1)(x - 2) , so f ′ (x) =0 at x=-1 and x=2

f ′ ′ (x) = 6x - 3 f ′ ′ (-1) = 6*(-1) – 3 = -9 < 0, so x = -1 is a local maximum

f ′ ′ (2) = 6*2 – 3 = 9 > 0, so x = 2 is a local minimum

Using f ′ ′ to find local extrema

First derivative:

y is positive Curve is rising.

y is negative Curve is falling.

y is zero Possible local maximum or minimum.

Second derivative:

y is positive Curve is concave up.

y is negative Curve is concave down.

y is zero Possible inflection point(where concavity changes).

Summary of what y ′ and y ′ ′ say about the curve

Example(cont.): Sketch the curve of f (x) = x3 – 1.5x2 – 6x + 5.

From previous slides,

f ′ (x) > 0 for x < -1 and x > 2 ; thus the curve is increasing on (-, -1) and (2, ) .

f ′ (x) < 0 for -1 < x < 2 ; thus the curve is decreasing on (-1, 2) .

f ′ ′ (x) > 0 for x > 0.5 ; thus the curve is concave upward on (0.5, ) .

f ′ ′ (x) < 0 for x < 0.5 ; thus the curve is concave downward on (-, 0.5)

(-1, 8.5) is a local maximum; (2, -5) is a local minimum.

(0.5, 1.75) is an inflection point.

(-1, 8.5)

(0.5, 1.75)

(2, - 5)

-1 2

Guidelines for sketching a curve:Step 1 : Study of Symmetry –Check for equation

f(x,y)= 0 is Symmetric about When equation is X-axis unaffected if y is changed to –y i.e. equation involves only even powers of y.

Y-axis unaffected if x is changed to –x i.e. equation involves only even powers x.

y=x unaffected if x & y are interchanged

y=-x unaffected if x & -y are interchanged

Oppo. unchanged if replace x by –x & y by -y Quadrants

Step 2 : Study behavior of curve near origin • If equation has no constant term then curve passes through

origin and then find the tangent at origin by equating the lowest degree terms in the equation to zero.

Decide whether O is a node, cusp or an isolated point

When there are two or more tangents at a point it is called a double point or multiple point.

If the tangents at a double point are • real and distinct then it is called a node.• real and coincident then it is called a cusp. • imaginary then it is an isolated points, which can not be

traced in the graph.

Step 3 : Find points of intersection with the axes• Find point of intersection with x-axis (if any) by putting

y=0 in the equation of the curve.

• Shift origin to this point and then using step 2 find

whether this point is a node, cusp or an isolated point.

• If the curve is symmetric about the line y = ± x then find

the point of intersection of the curve and the line y = ± x .

• Shift origin to this point and then using step 2 find

whether this point is a node, cusp or an isolated point.

Step 4 : Study of asymptotes

Vertical Asymptote :

Line of the form x = a, where a is a constant, for which f(x)-> or f(x)->- as x->a. Y-axis could be a vertical asymptote.

• Horizontal Asymptotes :

Lines of the form y = a, where a is a constant, for which f(x) -> a as f(x)-> or f(x)->-. X-axis could be a horizontal asymptote.

Oblique Asymptotes :

Lines of the form y = mx + b, where m is the slope and b is any constant, such that f(x) approaches mx + b as x-> or x->-.

• :• Step 5: Find horizontal extent or region The

horizontal extent is defined by those values of x for which y is defined. Find it from expression y = f(x). Similarly using x = g(y) find vertical extent of the curve. Eg. If y2 is negative for x > a then the curve does not lie to the right of ordinate of x = a.

• :• Step 6 : Study of special points Tangents parallel

to axes Solve y’=0 or x’= 0 to get critical points Solve y’=∞ at which curve changes character Points of extremum Maximum if y’ changes from + to – at x=c Minimum if y’ changes from - to + at x=c No extremum if y’ does not change in sign Points of inflection and concavity Concave up y” > 0 Concave down y” < 0

Trace the curve:

1.Symmetry :