The First Derivative Test

                                                                                                                 Briya Kirksey
We have now started to learn about the derivative application #3 which is Curve Sketching. You can find this section in our calculus book labeled as "Connecting the Graphs of f ', f '', and f''' ". The first part of this section is "The First Derivative Test".

                           The First Derivative Test 

     What are we looking for?

• We are taking a functions equation and determining where it is  increasing and/ or decreasing, and its relative extrema (relative maximum/minimum) on its graph. 

   

  How do we do this?

  1.   Take the derivative of the given function (if its not already given). 
  2.   Set the derivative equal to zero. Why? - Because that is where the slopes of the functions graph changes signs (positive to negative or vise versa). The values we get by doing this is what we know as critical points. 
  3. Make an x-axis number line labeling all of the critical points on it. 
  4. Choose numbers on the number before, after, and in between the critical points, known as testing points. Example: If critical points are x= -1, 0, 3 , then my testing points could be x= -1.5, -.5, 2, 4. 
  5. Plug each of the testing points into the derivative. Why?- To determine the behavior of the functions graph in between its critical points. 
  6. The answers produced by doing step number 5 will give you the x intervals in which the graph is increasing/decreasing and which x values there may be a max/min.  

                 Tip: 

  • When plugging in testing points into a functions derivative, pay attention to the values sign to make it easier in determining whether the slope will be positive or negative.
   Example:  y' = x (x-3)^2   testing point : -1
  
   Without doing any math, one can already tell that the slope will end up being negative. The value in the parenthesis is -4, but the fact that it is then squared makes it positive. However the x value  (-1) being multiplied to the positive value makes the slope  negative.   

  Concavity : Refers to the way Graphs look

  This is the worksheet and notes used to learn about Concavity. Concave up= a u-shape facing up. Concave down= u -shape facing downward. Inflation point= point on graph where cure changes; where y '' = 0. 

   

  

  
  

     

   

Existence Theorems

This week, we learned about the first application of derivatives: Existence Theorems. Basically, the theorems state that a certain point exists if it meets some criteria. Here are the four types of Existence Theorems we learned about.

1. Intermediate Value Theorem (IVT)
     - if a function f(x) is continuous on a closed interval [a,b]
     - and k (a y-value) is any value between f(a) and f(b)
     - then there is at least one number "c" (an x-value) in [a,b] such that f(c)= k

An example question: if f(x)= x^3+2x-1 is there a value on the interval [0,1] where c is 0?
     -First, determine if the function is continuous on the interval: in this case, it is
     -Second, find the values of f(0) and f(1)
          -      f(0)= -1      f(1)= 2
     -Since -1<0<2, a "c" exists on the interval such that f(c)= 0
     -To find this, we simply set the equation equal to 0.
          -      x^3+2x-1=0    if you solve this, and plug it into the calculator, c=.453   

2. Extreme Value Theorem (EVT)
     -if f(x) is continuous on the interval [a,b]
     -then f(x) is guaranteed to have an absolute minimum and an absolute maximum on the interval

An Extreme Value can happen at two places: endpoints and critical points (where the derivative is equal to 0 or is undefined)

An Example Question: Can you apply EVT to the equation x^-1 +4x on the interval [-4, -.25]? If so, what are the absolute maximum and minimum?
     -First, determine if the function is continuous on the interval.
          - x is not defined at 0, but this is not in the interval, so it doesn't matter
     -Second, evaluate f(-4) and f(-.25) to find possible points for extreme values
          - f(-4)= -16.25     f(-.25)= -5
     -Third, find the critical points by finding f'(x) and determining where the derivative= 0 or und.
          - f'(x)= -1    +4
                      x^2   
          - this is undefined at x=0, but isn't on the interval, so it doesn't matter
          - f'(x) equals zero at x= .5 or -.5 (however, .5 is not on our interval so it doesn't matter)
     -Fourth, evaluate f(x) at critical points
          - f(-.5)= -4
     -Finally, we can determine where the absolute maxes and mins are. In this situation, the maximum is -4 at x= -.5   and the minimum is -16.25 at x= -4

3. Mean Value Theorem (MVT)
     - if f(x) is continuous on [a,b] AND
     - if f(x) is differentiable on (a,b)
     - then there exists a number "c" in (a,b) such that
                 f'(c)= f(b)-f(a)
                              b-a

An Example Question: Is MVT applicable to the equation x+1/x on the interval [.5, 2]? If so, find the value of c.                                                                            
     - First, determine if the function is continuous and differentiable on the interval. In this case, it is, so MVT applies.
     - Second, evaluate f(.5) and f(2)
            -  f(.5)= 3    f(2)= 1.5
     - Third, plug these values into f(b)-f(a)
                                                          b-a
               In this case, it will simplify to -1
     - Fourth, find f'(x) and set it equal to -1.
          -   -1= -1  
                     x^2
          - x= 1, -1 (however, -1 is not on the interval, so we disregard this.) c=1

4. Rolle's Theorem (RT)
     - if f(x) is continuous on [a,b] AND
     - f(x) is differentiable on (a,b) AND
     - f(a)=f(b)
     - then there exists some number "c" on the interval (a,b) such that f'(c)=0

An Example Question: Is RT applicable to the equation f(x)=sin2x on the interval [π/2, 3π/8]? If so, find the value of c.
     - First, determine if the function is continuous and differentiable on the interval.
 In this case, it is.
     - Second, determine what f(π/2) is and f(3π/8) is.
          - f(π/2)= √2/2  f(3π/8)= √2/2 so yes, they are equal and MVT applies.
     - Third, find f'(x) and set it equal to 0 to find c.
          - f'(x)= 2cos2x 
               - 2cos2x= 0           x=π/4, 3π/4 (however, 3π/4 is not on the interval, so it is not included. c=π/4


That is what we did in class this week! Existence Theorems are so fun just like calculus yay!