CURVE SKETCHING 04/20/2015

CURVE SKETCHING

Hello Class this is our new class that will discuss Curve Sketching. In this class we will be talking about the application of differential calculus.

So, when we sketch a curve we look at several factors.

1- FIRST ORDER DERIVATIVE

X-intercept

We assume y=0 then solve for x.

Y-intercept

We assume x=0 then solve for y.

Find Maxima and Minimum.

Here these points have F’(x)= 0. This means that the first derivative of such a point is equal to zero. The maximum point is represented in the graph below by C while the minimum point is sown by B.

Increasing and decreasing function.

In the graph above, between point A and B the slope of the lines of tangent are negative thus the derivative F’(x) is negative. The function is therefore decreasing between A and B. it is also decreasing between C and D. Between B and C the slope of tangent lines is positive. The graph thus shows an increase between B and C.

Let F be continuous on the interval a<x<b and having derivative f on a<x<b and with a derivative F on a<x<b; then

If F’(x)>0 for a<x<b then F(x) will increase on a< x< b

If F’(x)<0 for a<x<b then F(x) is decreasing

If F’(x)=0 for a<x<b then F(x) is a constant on a<x<b

Up to that point is known as the First Derivative Test.

Example.

Use the First Derivative Test to solve F (x) = x³-27x

Solution.

      X-intercept

X (x²-27) = 0

Intercept at X=0, 5.2 and -5.2

      Y-intercept.

X(x²-27) =y

Y=0

      Maxima and Minima.

F’(x) = 3x²-27

We obtain the critical points when we equate F’(x) to zero.

3(x²-9) =0

(x² -9) =0

(x+3)(x-3)=0

X=-3 or 3

The curve has three sections

(-∞,-3)(-3,3)(3,∞)

This can be represented on the number line as below.

We will look at each interval to determine if it is increasing or decreasing.

In the interval (-∞,-3) we may use x= -5

F’(-5) = 3(-5)² -27 =75 -27 =48 >0

In the interval (-3,3) let x=0

F’(0) = 3(0) -27 = -27< 0

Maximum value F(-3)=(-3)³ – 27 (-3)= -27 + 81= 54

In the interval (3,∞) let x=5

F’(5)=3(5)² – 27= 48 >0

Minimum value F(3)=(3)³ – 27(3) =27 – 81 = -54

Interval	-∞,-3	-3,3	3,∞
F’(x)	positive	negative	positive
curve	Rise	Fall	Rise

2- SECOND ORDER.

We first find the x-intercept as in the first derivative.

From there the y-intercept can be found just like in the first derivative.

Symmetry

This can be found in three different ways

Replacing x by –x in the equation and the equation remains the same. This means it is symmetrical about the y axis.

Replacing y by –y then the sign of the equation changes. This means it is symmetrical about the x-axis.

We may also interchange y with x. this means symmetry about the line y=x

Asymptotes

Vertical asymptotes.

When the denominator of a rational function approaches zero, it has a vertical asymptote. Find the value of x that makes the denominator equal to zero. If your function is y = 1/(x+4), you would solve the equation x+4 = 0, which is x = -4. There may be more than one possible solution for more complex functions. Take the limit of the function as x approaches the value you found from both directions. For this example, as x approaches -4 from the left, y approaches negative infinity; when -4 is approached from the right, y approaches positive infinity. This means the graph of the function splits at the discontinuity, jumping from negative infinity to positive infinity.

Horizontal Asymptotes

The function is y = x/(x-1). Take the limit of the function as x approaches infinity. In this example, the "1" can be ignored because it becomes insignificant as x approaches infinity. Infinity minus 1 is still infinity. So, the function becomes x/x, which equals 1. Therefore, the limit as x approaches infinity of x/(x-1) = 1. Use the solution of the limit to write your asymptote equation. If the solution is a fixed value, there is a horizontal asymptote, but if the solution is infinity, there is no horizontal asymptote. If the solution is another function, there is an asymptote, but it is neither horizontal nor vertical

Maxima and minima

We can then find the maxima and minima of the curve as in the procedure for First Derivative Test.

Point of Inflection.

Let F(x) be differentiable two time [F”(x)] on a< x < b then

If F”(x)>0 for a<x<b then F(x) is Concave upwards.

If F”(x)<0 on a<x<b then F(x) is concave downwards.

If F”(x)=0 or not defined it is a point of inflection.

Example.

Draw the graph y= x⁴ -6x²

Solution

X -intercept y=0

0= x⁴ -6x²

X²(x²-6) = 0

X= 0, 6, -6

Y-intercept x=0

This is at origin.

Maxima and minima

y’ = 4x³ -12x

y’=4x(x+√3)(x-√3)

for maxima and minima put y’=0

then, 4x(x+√3)(x-√3)=0

x=0, √3, √-3

This function has its maxima and minima at (0,0), (-√3, -9) and (√3,9)

 y" = 12x² - 12
Now, y" = 0, then 12(x² - 1) = 0 ⇒ x = -1,1
For point of inflections, if x < -1, y" > 0 and -1 < x < 1, we have y" < 0. So, we get (-1,-5) is one of the point of inflection. Again, if x > 1, y" > 0, at this point again the sign of y" is changed. So, (1, -5) is the point of inflection.