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# Calculus Sections 4.1 and 4.3

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### Transcript

• 1. Calculus Sections 4.1 and 4.3
• Extreme Values of Functions
• Connecting f’ and f’’ the Graph f(x)
• 2. 4.1: Extreme Values of Functions
• Extreme Value Theorem: If f is continuous on a closed interval [a,b], than f has both a maximum value and a minimum value on the interval.
This graph is continuous because it is a closed interval between two points, meaning it will have both a maximum value and minimum value.
• 3. Absolute Extreme Values
• f is a function with a domain B, f(c) is the…
• * absolute maximum value if and only if
• for all of x in the domain B.
• * absolute minimum value if and only if
• for all of x in the domain B.
• 4. Finding Extreme Values
• Try: Find the absolute extreme Values
• 1. 2.
• 5. Finding Extreme Values **Solution**
• The graph is continuous on a closed interval [satisfying the Extreme Value Theorem], so:
• Graph 1 has a maximum @ x=c
• and
• minimum @ x= d
• 2. Graph 2 is not continuous and has an open interval. This does not satisfy the Extreme Value Theorem.
• Graph 2 has a
• maximum @ x= 1
• and
• no minimum.
• 6. Local Extreme Values
• Suppose c is an interior point of the domain of the function f . f(c) is a…
• * local maximum value at c if and only if for all x in some open interval containing c.
• * local maximum value at c if and only if
• For all x in some open interval containing c.
• If a function f has has a local max. or local min. value at an interior point c of its domain and f’ exists at c , then f’(c) = 0
• 7. Finding Extreme Values
• Try: Find the Local max and local min. values.
• 8. Finding Extreme Values **Solution**
• f(x) has a local maximum @
• x= x a and x c
• F(x) has a local minimum @ x= x b
• 9. Critical Points
• Definition: a point in the interior of the domain of a function f at which f’ = 0 of f’ does not exist.
• Critical points can be:
• Minimum
• Maximum
• Inflection points
• 10. 4.3: Connecting f’ and f’’ with the Graph of f
• When f’ changes from positive to negative on a graph,, than f will have a local max. at the value where it changes from positive to negative.
• When f’ changes from negative to positive on a graph,, than f will have a local max. at the value where it changes from negative to positive.
• If f’ doesn’t change signs anywhere on the graph, than there will be no local extreme value
• 11. Using “The Sign Line”
• When Given a problem: y = 2x 4 -4x 2 +1
• Find Critical Points
• Find the Derivative. [y’ = 8x 3 – 8x]
• Solve for x when y’=0 [8x 3 – 8x = 0]
• x = 0,1
• Do the sign line , plug in a number less and greater than the values of x:
• - + +
• (-1) (½) (2)
• 0 1
• 12. **Solution Continued**
• Therefore, there is
• a local maximum at x = 0,1
• a local minimum at x = 1,-1
• 13. Using the 2 nd Derivative To find Concavitiy
• Find critical numbers [f '(c) = 0 or f '(c) undefined] [y’ = 12x 2 + 42x + 36]
• Find f ''(x). Find f ''(c) for all critical numbers. [y’’ = 24x + 42]
• Set 2 nd derivative to 0 and solve for x.
• [0 = 24x + 42; x = -7/4]
• 4. Do the Sign Line
• -2 (-) (+)1
• -7/4
• 14. Concavity contd.
•   -If f ''(c) > 0, then f is concave up and f(c) is a relative min - f f ''(c) < 0, then f is concave down and f(c) is a relative max - If f ''(c) = 0, then the test fails.
• [Concave up from (-7/4, )
• Concave down from (- , -7/4)]
• 15. THE END!