"Exercise Lesson Comprehensive Application of Monotonicity and Parity" PPT on the Concept and Properties of Functions

"Exercise Lesson Comprehensive Application of Monotonicity and Parity" PPT on the Concept and Properties of Functions

Part One: Explanation of Curriculum Standards

1. Understand the relationship between parity and monotonicity of functions.

2. Be able to use the monotonicity and parity of functions to solve comprehensive problems such as comparing sizes, finding optimal values, and solving inequalities.

PPT on the comprehensive application of monotonicity and parity in exercise lessons, part 2: independent preview

Monotonicity of odd and even functions on symmetric intervals

1. (1) It is known that the function y=f(x) is an odd function on R and is an increasing function at (0,+∞). Then y=f(x) is in its symmetric interval (-∞,0 ) How monotonic is the above?

Tip: The graph of the odd function is symmetric about the coordinate origin, so the monotonicity is the same in the two symmetric intervals. That is, y=f(x) increases monotonically on its symmetric interval (-∞, 0).

(2) Can you prove the above conclusion using the definition of function monotonicity?

Tip: ∀x1,x2∈(-∞,0), and x1-x2>0,

∵y=f(x) is an increasing function on (0,+∞),

∴f(-x1)>f(-x2).

∵y=f(x) is an odd function on R,

∴f(-x1)=-f(x1),f(-x2)=-f(x2),

∴-f(x1)>-f(x2), ∴f(x1)

∴The function y=f(x) is an increasing function on (0,+∞).

(3) It is known that the function y=f(x) is an even function on R and a decreasing function at (0, +∞). y=f(x) on its symmetric interval (-∞, 0) is Increasing function or decreasing function?

Tip: The graph of the even function is symmetric about the y-axis, so the monotonicity is opposite in the two symmetric intervals. That is, y=f(x) increases monotonically on its symmetric interval (-∞, 0).

(4) Can you prove the above conclusion using the definition of function monotonicity?

Tip: ∀x1,x2∈(-∞,0), and x1-x2>0,

∵y=f(x) is a decreasing function on (0,+∞),

∴f(-x1)

∵y=f(x) is an even function on R,

∴f(-x1)=f(x1),f(-x2)=f(x2),

∴f(x1)

∴The function y=f(x) is an increasing function on (0,+∞).

2. Fill in the blanks

(1) If the function f(x) is an odd function, and f(x) is a monotonic function on the interval [a,b], then f(x) is on its symmetric interval [-b , -a] is also monotonic, and the monotonicity is the same.

(2) If the function f(x) is an even function, and f(x) is a monotonic function on the interval [a,b], then f(x) is on its symmetric interval ["− " ?",−" ?] is also monotonic, and the monotonicity is opposite.

3. Do it

(1) If the odd function f(x) is a decreasing function on [-6,-2], and the minimum value is 1, then it is ( )

A. Increasing function and the minimum value is -1 B. Increasing function and the maximum value is -1

C. Decrease the function and the maximum value is -1 D. Decrease the function and the minimum value is -1

Analysis: ∵ odd function f(x) is a decreasing function on [-6,-2], and the minimum value is 1, ∴ function f(x) is on [2,6] above is a decreasing function and the maximum value is -1.

Answer:C

(2) If the even function f(x) is an increasing function on (-∞,0], then the magnitude relationship of f(-5), f( ), f(-2), f(4) is_ ____________________.

Analysis: Because f(x) is an even function and is an increasing function on (-∞,0], so f(x) is a decreasing function on [0,+∞), and f(- 5)=f(5),f(-2)=f(2). Because √3<2<4<5, so f(5)

Comprehensive application of monotonicity and parity in exercise lessons PPT, part three: inquiry learning

Apply the monotonicity and parity of the function to determine the size of the function value

Example 1 It is known that the domain of the even function f(x) is R. When x∈[0,+∞), f(x) is an increasing function, then f(-2), f(π), The magnitude relationship of f(-3) is ()

A.f(π)>f(-3)>f(-2)

B.f(π)>f(-2)>f(-3)

C.f(π)

D.f(π)

Analysis: ∵f(x) is an even function on R, ∴f(-2)=f(2), f(-3)=f(3).∵2<3<π, and f(x) is The interval [0,+∞) is an increasing function, ∴f(2)

∴f(-2)

Answer:A

Reflect on the application of the monotonicity and parity of the function when judging the size of the function value. First, use the parity of the function to transform the independent variables into the same monotonic interval, and then compare the size of the function value based on the monotonicity of the function.

Extended exploration (1) If the "increasing function" in this example is changed to "decreasing function", and other conditions remain unchanged, what is the relationship between f(-2), f(π), and f(-3)?

(2) If the "even function" in this example is changed to an "odd function", and other conditions remain unchanged, compare the sizes of these three numbers.

Solution: (1) Because when x∈[0,+∞), f(x) is a decreasing function, so f(2)>f(3)>f(π). And because f(x ) is an even function on R, so f(-2)=f(2), f(-3)=f(3), thus f(-2)>f(-3)>f(π).

(2) Because the function is an odd function defined on R and is an increasing function on [0,+∞), the function is an increasing function on R,

Because -3<-2<π, so f(-3)

Comprehensive application of monotonicity and parity in exercise lessons PPT, part 4: thinking analysis

Determine parity of abstract functions

Typical example: Given functions f(x), x∈R, if for any real number a, b, f(a+b)=f(a)+f(b), verify: function f(x) is an odd function .

Proof: It can be seen from the meaning of the question that the domain of the function is R and is symmetric about the origin.

Let a=0, then f(b)=f(0)+f(b), ∴f(0)=0.

Let a=-x, b=x and substitute, we get f(-x+x)=f(-x)+f(x), that is, 0=f(-x)+f(x),∴f( -x)=-f(x),

∴The function f(x) is an odd function.

Reflection and insights: To judge the parity of an abstract function, we mainly use the assignment method and combine it with known conditions to find the relationship between f(-x) and f(x) to draw conclusions.

Variant training is a known function f(x), x∈R. If for any real number x1, x2, there is f(x1+x2)+f(x1-x2)=2f(x1)·f(x2),

Prove: The function f(x) is an even function.

Proof: Let x1=0, x2=x, get f(x)+f(-x)=2f(0)f(x).①

Let x2=0, x1=x, get f(x)+f(x)=2f(0)f(x).②

From ①②, we get f(x)+f(-x)=f(x)+f(x), that is, f(-x)=f(x),

So the function f(x) is an even function.

Comprehensive application of monotonicity and parity in exercise lessons PPT, part 5: in-class practice

1. If f(x) is an even function defined on [-6,6], and f(4)>f(1), then the following formulas must be true ()

A.f(0)f(3) C.f(2)>f(0) D.f(-1)

Analysis: ∵f(x) is an even function defined on [-6,6],

∴f(-1)=f(1). And f(4)>f(1), f(4)>f(-1).

Answer:D

2. If f(x) satisfies f(-x)=f(x), and f(x) is an increasing function on (-∞,-1], then ()

A.f("-" 3/2)

B.f(-1)

C.f(2)

D.f(2)

Analysis:∵f(-x)=f(x),∴f(2)=f(-2),

∵-2<-3/2<-1, and f(x) is an increasing function on (-∞,-1], ∴f(-2)

Answer:D

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