Math problems, math problems about functions, math problems about functions

So 2f(m)=2f(m)f(0), so f(0)=1, so f(0)=1

2.Let x, substitute y, so f(0+x)+f(o-x)=2f(0)f(-x).

becausef(0)=1, so f(x)=f(-x), so f(x) is an even function.

Typing is too cumbersome ,......

3.Let x=x+c 2 y=c 2,sof(x+c)+f(x)=2f(x+c 2)(c 2), because f(c 2)=o,so f(x+c)=-f(x).

1.Let x=0, y=0

f(0)+f(0)=2f(0)f(0)=2f(0)Since f(0)≠0, then f(0)=1

2.Let x=0

then f(y)+f(-y)=2f(0)f(y)=2f(y), then f(y)=f(-y).

y=f(x) is an even function.

Let x=x+c 2, y=c 2

f(x+c)+f(x)=2f(x+c 2)f(c 2)=0, then f(x+c)=-f(x).

Solution: (1) B; The water is submerged in the iron;

2) The analytic formulas of the line segments ab and de are as follows: y1=k1x+b, y2=k2x+b, ab passes through points (0,2) and (4,14), and dc passes through (0,12) and (6,0).

4k1+b1=1

4b1=2b2=12

6k2+b2=0

Solve {k1=3

b1=2k2=-2

The analytic formula for b2=12 is y=3x+2 and y=-2x+12, so that 3x+2=-2x+12 is solved to x=2, when 2 minutes is the same height as the water surface of the two sinks

3) From the image, when the water surface is not covered by the iron block, the water surface rises by 12cm in 4 minutes, that is, it rises by 3cm in 1 minute, when the water surface does not pass the iron block, it rises by 5cm in 2 minutes, that is, it rises in 1 minute, and the bottom area of the iron block is xcm, then 3 (36-x)=, the solution is x=6, and the volume of the iron block is: 6 14=84cm 3

4）（36×19-112）÷12=60cm^2

1.Second; Armor; At 4 minutes, the depth of water in tank B is 14 centimeters per minute (look at the figure to find the equation of the line where ab,de is located, and solve the system of equations), cubic centimeters (find the bottom area of groove A from the BC segment, and then calculate it according to the image of the first 4 minutes) square centimeters (calculated from the constant volume growth rate).

1) Function f(1) = -a+1, f'(x）=1/x-a，∴f 'The equation for the tangent l l at the point p(1,f(1)) at the point p(1,f(1)) is .

y-(-a+1)=(1-a)(x-1), i.e. y=(1-a)x

Find the difference between the tangent l and the function f(x)=lnx-ax+1, and get g(x)=(1-a)x,-(lnx-ax+1,).

g'(x)=(1-a),-1 x+a=1-1 x,,, easy to get g'(1) = 0, and x at (0,1) g'(x) <0, x at (1, ) g'(x) > 0, g(x) has a maximum value g(1)=0 at x=1, i.e., the image of the function y=f(x) (x is not equal to 1) is below the line l.

2) by f(x)=lnx-ax+1=0.It is easy to get the images of lnx=ax-1, and y=lnx and y==ax-1, depending on the value of a, the two images may intersect at two points, tangent to one point, or not intersect, corresponding to the zero point of f(x)=lnx-ax+1.

It's too cumbersome to make drawings and typing here, so please do it yourself.

That's what I thought.

1） f '(x) =1 x-a substitutes (1,f(1)) points into and solves f'(x) =1-a f(x)= a+1 The proof image after solving the tangent equation using the point oblique equation is below, as long as it is proved that the tangent equation is always greater than the equation of equal functions.

2) Change the function to lnx=ax-1 so that g=lnx and h=ax-1 and their number of intersections is the number of zeros

3) I don't know if you can ask for details (I forgot high school math), but please forgive me.

1.Derivative functions (which may be discussed in a categorical manner).

2.Special point method.

Decreasing by baiy with the increase of x yields du:

m-3<0.

zhi m<3

2) Let x=0 get y=-n+4

then -n+4<0

n>43) from the image over the DAO origin (get:

n+4=0, then n=4

Y is a function again.

So back to M-3≠0

m≠ （x≤100）

y=57+(x-100)* x>100)2) January, y1=76, 76=57+(x1-100)*

x1=138

In February, y2=63

63=57+（x2-100)*

x2=112

March, order y3=

x3 = 80 so the total electricity consumption = x1 + x2 + x3 = 330 kWh.

1。When m is greater than 3, because it is a proportional function.

2。When n is greater than 4, because the intersection with y, i.e. x is equal to 0, then the inner is less than 0

3.When n is equal to 4, allow m to be any real number.

4.When x is less than or equal to 100, y=

When x is greater than 100, y=100x+(x-100) multiplied.

f(x) is an even function defined on r, and when x 0, f(x) is a monotonic function, x=(x+3) (x+4), or -x=(x+3) (x+4), i.e. x +3x-3=0, or x +5x+3=0, from the relationship between the root and the coefficient, we can see that x +3x-3=0, and the sum of all roots of x +5x+3=0 = (-3) + (5) = -8

The type of question that combines trigonometric functions and fish derivatives.

sin2x=2sinxcosx,sin2x+cos2x= 2sin(x+) are simplified according to these two formulas.

Hope that helps!

(1) Solution: Let ** be x, and the monthly sales volume y.

According to the title, it is called:

y=[(80-x)÷2]×100

Simplify, get: y=50000-500x

2) Set the selling price to A

According to the title, it is called:

a-40)(50000-500a)

5(a-70)(a-70)-900

So: the maximum value is 70

A: The sales price is 70 when the income is the most.

Let y be the monthly sales volume and x be the product**.

y=50000-500x

Let x be the sales ** and y be the revenue.

y=(50000-500x)x-40(50000-500x)=-500(x-30)*2+250000

So when x=30, y is the maximum, which is 250000

Monotically decreasing interval: (-2, -1)u(0, positive infinity)f'（x）=2（1＋x）－[2/(1+x)]=2[(1+x)-1/(1+x)]>0

The time function increases monotonically, i.e., (1+x)>[1 (1+x)], and the solution is: -2