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Critical Points
Points where the derivative of a function equals zero. Critical values are where there is a minimum or maximum on the original function.
Example:
f(x)= x^2 – 4x
f’(x)= 2x – 4
2x – 4= 0
2x=4
x= 2 ← critical value
Rolles Theorem
Let f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one number c in (a,b) such that f'(c)=0. Quick Version - if endpoints are equal (y-values) then there is a point where there is a horizontal tangent (slope=0) |
Example 1. x(x-3)
Example 2. f(x)=ln(x-2) [3,5]
Example 3) f(x)=3sin(2x) [0,pi]
Mean Value Theorem
- If f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that
b - a
There is a number c that a<c<b
Steps to solve Mean Value Theorem
- Check to make sure f(x) is continuous
- Find the derivative of f(x)
- Find if the derivative is continuous
- Check to see if f(x) satisfies the two conditions for Mean Value Theorem
- Evaluate f(a) from the interval
- Evaluate f(b) from the interval
- Solve for x
- f(b)-f(a) = f ‘(c)(b-a)
- f’(c)= f(b)-f(a)
What the Mean Value Theorem tells us is that if we found the slope of the endpoints, then there is one place where the tangent line equals the slope.
Examples:
Example 1)
Example 2)
Example 3)
Concavity
1. Take the 2nd derivative and find the critical values at the second derivative .
2. Set up a chart plugging into the second derivative. Test for the sign (negative/ positive)
3. Set up a number line and identify where it's increasing and decreasing.
1. Take the 2nd derivative and find the critical values at the second derivative .
2. Set up a chart plugging into the second derivative. Test for the sign (negative/ positive)
3. Set up a number line and identify where it's increasing and decreasing.
Example 1:
Suppose f(x)=x^3-3x^2+x-2. Let's determine where the graph of f is concave up and where it is concave down. Since f is twice-differentiable for all X, we use the reslut given above and first determine that f"(x)=6(x-1). thus f"(X)>0 if x>1 and f"(x) <0 if x<0. By the concavity theorem, the graph of f is concave up for x>1 and concave down for x<1.
Suppose f(x)=x^3-3x^2+x-2. Let's determine where the graph of f is concave up and where it is concave down. Since f is twice-differentiable for all X, we use the reslut given above and first determine that f"(x)=6(x-1). thus f"(X)>0 if x>1 and f"(x) <0 if x<0. By the concavity theorem, the graph of f is concave up for x>1 and concave down for x<1.
Absolute Extrema
The largest and smallest value of a function on an interval.
Extreme Value Theorem: Suppose that f is continuous on the interval [a, b] then there are two numbers a ≤ c, d ≤ b so that f(c) is an absolute maximum for the function and f(d) is an absolute minimum for the function.
To find absolute extrema:
Examples:
The largest and smallest value of a function on an interval.
Extreme Value Theorem: Suppose that f is continuous on the interval [a, b] then there are two numbers a ≤ c, d ≤ b so that f(c) is an absolute maximum for the function and f(d) is an absolute minimum for the function.
To find absolute extrema:
- Find all critical values for the function f on the open interval.
- Evaluate each critical value on the closed interval in the function f.
- The largest function value from steps 2 and 3 is the absolute maximum and the smallest function value from steps 2 and 3 is the absolute minimum.
Examples:
Example 1)
Example 2)
Example 3)
Increasing and Decreasing
When the slope is becoming increasingly negative, the derivative is decreasing
When the slope is becoming increasingly positive, the derivative is increasing
How to find:
When the slope is becoming increasingly negative, the derivative is decreasing
When the slope is becoming increasingly positive, the derivative is increasing
How to find:
- Take the derivative
- Find the critical values
- Make a chart with the intervals
- Pick a test value within the intervals and plug into the derivative
- Check to see if it’s positive or negative
- Set up a number line with critical values and endpoints
- Find intervals and label increasing decreasing
For the examples please click on the link below.
https://www.educreations.com/lesson/view/increasing-and-decreasing/44836468/?s=KEEabo
https://www.educreations.com/lesson/view/increasing-and-decreasing/44836468/?s=KEEabo
Relative Extrema
- First Derivative Test
Each top of the curve are relative maximums and each bottom are relative minimums
A relative maximum is a point higher than the points around it and the relative minimum is a point lower than the points around it.
How to find relative extremes using First Derivative Test
Examples:
- First Derivative Test
Each top of the curve are relative maximums and each bottom are relative minimums
A relative maximum is a point higher than the points around it and the relative minimum is a point lower than the points around it.
How to find relative extremes using First Derivative Test
- Same steps as increasing/decreasing
- Identify relative extremes on number line
Examples:
1.
2.
3.
Second Derivative Test
Suppose that c is a critical point where f'(c)=0 that f'(x) exists in a neighborhood of c, and that f"(c) exists. Then f has a relative maximum value at c if f"(c)<0 and a relative minimum value at c if f"(c)>0. If f"(c)=0, the test is not informative.
Suppose that c is a critical point where f'(c)=0 that f'(x) exists in a neighborhood of c, and that f"(c) exists. Then f has a relative maximum value at c if f"(c)<0 and a relative minimum value at c if f"(c)>0. If f"(c)=0, the test is not informative.