A question about monotonicity of function

A function monotonicity question

1))g(x)=x has two unequal real roots

(bx-1)/(a^2x 2b)=x

b^2- 4a^2>0

The absolute value of b > the absolute value of 2a

When a>0, b>2a

f(x) The image opening is upward, the axis of symmetry x= - b/2a

So f(x) is an increasing function at (-1, positive infinity)

So f(x) is an increasing function at (-1, 1)

When a

f(x) The image opening is downward, the symmetry axis x= -b/2a >1

So f(x) is an increasing function at (negative infinity, 1,)

So f(x) is an increasing function at (-1, 1)

To sum up, f(x) is a monotonically increasing function on (-1,1)

2.x3

a root (b^2-4a)>root (b^2-4a^2)>-root (b^2-4a^2)>-a root (b^2-4a).

It can be seen that a>0 is required, then a^2(b^2-4a)>b^2-4a^2.

(a-1)[b^2(a 1)-4a^2]>0 .

a>1, or a0).

So, a>1

Function monotonicity practice

1. Suppose y=f(x) is a decreasing function on R, and the monotonically decreasing interval of y=f(IX-3I)

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Suppose the function u=IX-3I, x∈R, which decreases monotonically on (-∞, 3], then y=f(u)=f(IX-3I) monotonically decreases on (-∞, 3] Increment;

The function u=IX-3I, x∈R, which increases monotonically on [3, ∞), then y=f(u)=f(IX-3I) decreases monotonically on [3, ∞);

That is, the monotonically decreasing interval of function y=f(IX-3I) is [3,∞)

-------------If you don’t understand, let’s put it another way:

x1│x2-3│, f(│x1-3│)

When 3

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It is known that the quadratic function f(x) satisfies f(0)=1, f(x 1)-f(x)=2x, try the analytical formula of f(x)

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Let’s assume the quadratic function f(x)=ax^2 bx c

From f(0)=1, we get c=1

So, f(x)=ax^2 bx 1

So f(x 1)=a(x 1)^2 b(x 1) 1

f(x)=ax^2 bx 1

So f(x 1)-f(x)=2ax a b

It is known that f(x 1)-f(x)=2x

Then the polynomial 2ax a b about x is equal to 2x, and its coefficients are equal

Therefore, a=1, and a b=0, then b=-1

f(x)=x^2-x 1

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2. It is known that the function f(x) defined on [1,4] is a decreasing function and a set of real numbers a that satisfies the inequality f(1-2a)-f(4 a)>0

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Change the inequality to f(1-2a)>f(4 a), and use the monotonicity of the function to get rid of the corresponding rule f, pay attention to the domain of the function

The domain of function f(x) is [1,4], and it is a subtractive function. Then the real number a satisfies the following three inequalities at the same time:

1

1

1-2a

Solving the inequality group, we get: -1

So, the value range of the real number a is (-1,0]

Compared with question 2, please do question 3 yourself......

Ask a question about quadratic functions and monotonicity

1) Analysis: ∵The axis of symmetry is the quadratic function y=f(x) of X=-1. The minimum value on R is 0, and f(1)=1

Suppose the function f(x)=ax^2 bx c=a(x b/(2a))^2 (4ac-b^2)/4a

∴a>0,-b/(2a)=-1==>b=2a,(4ac-b^2)/4a=0==>4ac=b^2

∴4ac=4a^2==>c=a

Also a b c=1==>4a=1==>a=1/4,b=1/2,c=1/4

The analytical formula of the ∴ function is f(x)=1/4x^2 1/2x 1/4

2) If g(x)=(z 1)f(z-1)-zx-3 is an increasing function on X belonging to [-1,1], the value range of real number z

Analysis: 1)f(x)=1/4x^2 1/2x 1/4

f(x-1)=1/4x^2-1/2x 1/4 1/2x-1/2 1/4=1/4x^2

g(x)=(z 1)1/4x^2-zx-3=(z 1)/4{[x-2z/(z 1)]^2-[(4z^2 12z 12) /(z 1)^2]}

=(z 1)/4[x-2z/(z 1)]^2-(z^2 3z 3)/(z 1)

∵g(x) is an increasing function when X belongs to [-1,1]

When (z 1)/4>0==>z>-1

∴2z/(z 1)2zz

∴-1

When (z 1)/4z

∴2z/(z 1)>=1==>2zz>=1, obviously contradicts z

When (z 1)/4=0==>z=-1

∴g(x)=x-3, obviously g(x) is an increasing function when X belongs to [-1,1]

To sum up, it satisfies that g(x) is an increasing function when X belongs to [-1,1], -1

3) The largest real number m (m is greater than 1), such that there is a real number t. As long as X belongs to [1, m], f(x t) is less than or equal to x.

Analysis: 1)f(x)=1/4x^2 1/2x 1/4

f(x t)=1/4(x t 1)^2

(x t 1)^2

x^2 2(t-1)x (t 1)^2

When t=0, x^2-2x 1x=1

When t>0, ⊿=4(t-1)^2-4(t 1)^2=-16t

whent0

x1=(1-t)-2√（-t）, x2=(1-t) 2√（-t）

Let (1-t) 2√(-t)=1==>t=-4

∴m=x2=(1-t) 2√(-t)=9

∴There is a real number t=-4. As long as X belongs to [1,9], it is true that f(x-4t) is less than or equal to x.

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