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!
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