High School Function Problem Solving, Solving High School Function Math Problems

Let any x10, f(x2-x1)>1 i.e. f(x2)+f(-x1)-1>1,f(x2)+f(-x1)>2

If the middle order n=0 f(m)=f(m)+f(0)-1 is known, and f(0)=1 is known, then f(0)=f(m)+f(-m)-1f(-m)=2-f(m) f(x2)+2-f(x1)>2 f(x1)f(x) is an increasing function on r.

2) f(3) = f(2) + f(1)-1 f(2) = 2f(1)-1f(3) = 3f(1)-2=4 f(1)=2f(a2+a-5)<2=f(1) according to the monotonic increase in (1).

a 2+a-5<1 solution -3

What's this mess?

1) Let the arbitrary function m n belong to r, and n > 0f(n)-1 > 0 (when x is greater than 0, there is always f(x) greater than 1) m + n > m

f(m+n)=f(m)+f(n)-1

f(m+n) -f(m) =f(n)-1 >0, so it is an increasing function.

2)∵f(3)=f(2)+f(1)-1 ,f(2)=f(1)+f(1)-1=2f(1)-1

f(3)=3f(1)-2=4

f(1)=2

f(a2+a-5)<2=f(1) is monotonically increased according to (1).

a 2+a-5<1 solution.

The first derivative of f(x) is equal to x 2+2ax+b

Substituting the condition of knowing the potato age into this formula gives a number of traces, 1-2a+b=0, so b=2a-1

The first derivative of f(x) is (x-a) 2-(a-1) 2, which is a parabolic function and the axis of symmetry is x=-a, so it decreases monotonically in the interval (-a) and increases monotonically in the interval (-a,+ states).

f(-1)=1/3-(-1)^2=-2/3<0f(0)=3^0-0^2=1>0

and f(x) is a continuous function.

Therefore f(x)=0 has a real root on [-1,0].

Happy studying.

Because the function is continuous (not emphasized in high school), the function f(0)*f(1)<0 (zero point theorem) has a zero point, so there is a real root.

The definition of the odd function is f(x)=-f(-x), and since f(2)=6, f(-2)=-6, brought in, the solution is a=5.

1. f(x+1)=ax +(2a+b)x+a is an even function, and f(-x)=f(x). Therefore, 2a+b=0;The image of the function f(x) is tangent to the line y=x, ie.

ax +bx=x has and only one solution. b=1,a=-1 2;f(x)=-1/2x²+x。

2. If the constant k 2 3, there is an interval [m,n](m, this question is actually asking the question of the intersection of y=kx and f(x). The image of f(x) is known, and we can draw a sketch: (1) the line and the function have only one intersection point (i.e., (0,0)), and we use the line x=m,x=n to intercept it; We will find that the projection of the two sets of lines on the y-axis (the value of the function corresponding to the truncated lines) does not coincide, and the condition "such that the range of f(x) on the interval [m,n] is exactly [km,kn]" is not satisfied.

2) There are two intersection points between the line and the function, and obviously one of the intersection points is (0,0). Similarly, if we take two straight lines and cut them off, we find that when the other intersection of the line and the function is to the right of the line x=1 (the axis of symmetry of the quadratic function), the condition is not satisfied (the projections do not coincide).

This results in the other intersection to the left of the line x=1. k 2 3, then the other intersection is to the left of the line x=2 3. (1) If the other point is to the right of the y-axis (x=0), then the interval that satisfies the condition is [0,n], where n satisfies the condition, {n|0(2) The other point is to the left of the y-axis (x=0), then the interval that satisfies the condition is [m,0], where m satisfies the condition {m|m<0｝。

There are many descriptions, and of course, when writing, you can draw conclusions directly without using too much analysis.

(1) f(x+1)=ax +2ax+a+bx is an even function =>2a+b=0

The image of the function f(x) is tangent to the line y=x =>x=ax +bx, and there is only one solution =>b=1

a=-1/2

f(x)=-x²/2+x

First, substituting the p point to the original function equation gives b=2

Then find the derivative at x=1, f'(x)=2x 2-2ax-9, substitute x=1 into -12=2-2a-9

Find a=5 2

(1) Because the function is a quadratic function and a is a negative value, the function opening is downward, so there is a maximum value, and after factoring the function, we get y=-2(x-1) 2-1, so when x=1, the function value is the maximum: -1

2) This function can be regarded as a quadratic function to solve: function a is a positive opening upward, there is a minimum value, the original test is written as: y=x 2(3x 2-4x-12)+18, and 3x 2-4x-12 is factored to obtain 3(x-2 3) 2-10 9

Then the function has a minimum value when x=2 3: 1418 81

(1) y=-(2x 2-4x+3) Split item supplement allocation method =-(2x 2-4x+4-1).

Let x=1, get ymax=1

I didn't understand the meaning of =18 in the second question.

As for the second question, you can find the derivative.

12x^3-12x^2-24x

12x(x^2-x-2)

12x(x-2)(x+1)

After that, let x=0,2,-1

Extreme values are derived.

(1) The quadratic term coefficient of the quadratic function is negative, so the image opening is downward, and the symmetry axis is found - (b 2a), that is, x=1

So when x=1 is y there is a maximum, ymax=-1

The second method is to find a derivative.

y= -2x^2+4x-3;

y'=-4x+4 require'=0 -4x+4=0 x=1x （-1） 1 （1，+∞y'+ 0 -y increment maximum decreasing When x = 1, y has a maximum.

(1) Solution: The original formula is: y=-2(x-2) 2-1 When x=2, y takes the minimum value.

i.e.: y -1

Note: Formulas are used for such extreme value problems.