"End of Chapter Integration" Concept and Properties of Functions PPT

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"End of Chapter Integration" Concept and Properties of Functions PPT

Part One: Topic 1: Finding the Range of a Function

Example 1 Find the range of the following functions:

(1)y=(5x"-" 1)/(4x+2);(2)y=(x^2 "-" 4x+3)/(2x^2 "-" x"-" 1);

(3)y=(2x^2+4x"-" 7)/(x^2+2x+3);(4)y=2x-√(x"-" 1).

Solution: (1) (Solved with the help of the characteristics of the inverse proportional function)

y=(5x"-" 1)/(4x+2)=(5/4 "(" 4x+2")-" 1"-" 5/2)/(4x+2)=(5/4 " (" 4x+2")-" 7/2)/(4x+2)=5/4-7/(2"(" 4x+2")" ).

∵7/(2"(" 4x+2")" )≠0,∴y≠5/4.

So the value range of the function is {y"∈" R├|y≠5/4┤}.

(2)∵y=(x^2 "-" 4x+3)/(2x^2 "-" x"-" 1)=("(" x"-" 1")(" x"-" 3 ")" )/("(" x"-" 1")(" 2x+1")" )=(x"-" 3)/(2x+1)(x≠1),

And (x"-" 3)/(2x+1)=(1/2 "(" 2x+1")-" 7/2)/(2x+1)=1/2-7/(2"( " 2x+1")" ).

When x=1, the original formula y=(1"-" 3)/(2×1+1)=-2/3.

The value range of the ∴ function is {y"∈" R├|y≠1/2 "and" y≠"-" 3/2┤}.

(3) (Convert into a quadratic equation about x, and then use the discriminant to evaluate the domain)

It is known that the functional formula can be transformed into: yx2+2yx+3y=2x2+4x-7.

(y-2)x2+2(y-2)x+3y+7=0,

When y≠2, consider the above equation as a quadratic equation about x.

∵x∈R,∴Δ≥0, that is, [2(y-2)]2-4(y-2)(3y+7)≥0.

The solution is -9/2≤y<2.

When y=2, 3×2+7≠0, ∴y≠2.

The value range of the ∴ function is -9/2,2.

(4) Let √(x"-" 1)=t, then t≥0, x=t2+1.

∴y=2(t2+1)-t=2t2-t+2=2 t-1/4 2+15/8.

∵t≥0,∴y≥15/8. The value range of ∴ function y=2x-√(x"-" 1) is 15/8,+∞.

Integrate PPT at the end of the chapter, the second part of the content: Topic 2: Using the monotonicity of the function to find the maximum value of the function

Example 2 Let a be a real number, function f(x)=x2+x-a+1, x∈R.

(1) Discuss the parity of function f(x);

(2) Find the minimum value of f(x).

Solution: (1) When a=0, function f(-x)=(-x)2+|-x|+1=f(x), at this time f(x) is an even function .

When a≠0, f(a)=a2+1,f(-a)=a2+2|a|+1,f(-a)≠f(a),f(-a) ≠-f(a).

At this time, the function f(x) is neither an odd function nor an even function.

(2)①When x≤a, function f(x)=x2-x+a+1= x-1/2 2+a+3/4.

If a≤1/2, then the function f(x) decreases monotonically on (-∞,a], so the minimum value of the function f(x) on (-∞,a] is f(a )=a2+1.

If a>1/2, then the minimum value of function f(x) on (-∞,a] is f 1/2 =3/4+a, and f 1/2

②When x≥a, function f(x)=x2+x-a+1= x+1/2 2-a+3/4.

If a≤-1/2, then the minimum value of function f(x) on [a,+∞) is f -1/2 =3/4-a, and f -1/2 ≤f( a).

If a>-1/2, then the function f(x) increases monotonically on [a,+∞), so the minimum value of the function f(x) on [a,+∞) is f (a)=a2+1.

To sum up, when a≤-1/2, the minimum value of function f(x) is 3/4-a;

When -1/2

When a>1/2, the minimum value of function f(x) is a+3/4.

Methods and techniques The basic idea of ​​solving problems with parameters is to classify and discuss. The key is to determine the standard of discussion, which requires no repetition or omission. The standard for discussion of parity in this question is zero and non-zero parameters, which correspond to even functions and non-odd functions respectively. Even functions; the discussion standards for maximum and minimum values ​​are relatively complicated and can be seen as two types of standards. One is the zero point of the absolute value (zero point knowledge will be learned in Chapter 4), and the other is the symmetry axis of the parabola and the corresponding interval. Position usually requires the help of the image of the function.

Integrate PPT at the end of the chapter, the third part: Topic 3 Application of parity of functions

Example 3 If the odd function y=f(x) is a decreasing function defined on [-1,1], and f(1-a)+f(1-a2)>0, find the Ranges.

Solution: According to the properties of odd functions, -f(1-a2)=f(a2-1), that is, f(1-a)+f(1-a2)>0 is equivalent to f(1-a)>f (a2-1),

And because f(x) is a decreasing function defined on [-1,1],

So {■("-" 1≤1"-" a≤1"," @"-" 1≤a^2 "-" 1≤1"," @1"-" a

Methods and techniques: Use the property that f(x) is an odd function and a decreasing function, remove f, and equivalently transform the inequality set of a.

Variation Training 3 If f(x) is an even function defined on the real number set R and is an increasing function on the interval (-∞,0), and f(2a2+a+1)

Solution: Method 1: ∀x1,x2∈(0,+∞), and x1-x2,

Because f(x) is an increasing function on the interval (-∞,0),

So f(-x1)>f(-x2).

And because f(x) is an even function, f(x1)>f(x2),

So f(x) is a decreasing function on (0,+∞),

Because 2a2+a+1=2 a2+1/2a +1=2 a+1/4 2+7/8,3a2-2a+1=3 a-1/3 2+2/3,

So 2a2+a+1 and 3a2-2a+1 are two positive numbers,

Therefore, f(2a2+a+1)3a2-2a+1, and the solution is 0

Method 2: Same as Method 1, it is judged that 2a2+a+1 and 3a2-2a+1 are two positive numbers, then -(2a2+a+1)<0 and -(3a2-2a+1)<0.

According to the properties of even functions, f(2a2+a+1)

In addition, f(x) is an increasing function on the interval (-∞,0), that is, -(2a2+a+1)<-(3a2-2a+1), and the solution is 0

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