Proof of function question 50, a proof of function question?

If the function f(x) satisfies f(x1+x2)=f(x1)·f(x2),(x r)and it is known that f(x) is a monotonic function on r.

Verification: f(x) is an exponential function.

Proof: 1).Let x1=x2=0, then f(0)=f2(0) gives f(0)=0 or f(0)=1

When f(0)=0, let x1=x, x2=-xthen f(0)=f(x)f(-x)=0

Yes, you can get f(x) or f(-x)=0, because x is arbitrarily selected. There is f(x) 0It contradicts that f(x) is a monotonic function on r, so f(0)=1

2).Let x1=x2=x, then f(2x)=f 2(x). Derive it.

f'(2x)=f(x)f'(x) divided by f(2x) = f 2(x).

f'(2x)/f(2x)=f'(x) f(x), which is equal to a constant c because it is true for any x. Namely.

f'(x)/f(x)=c

The process of solving the solution is slightly as follows:

dlnf(x)=cdx

lnf(x)=cx+d d is the integration constant.

f(x)=d'e^(cx)=d'(e^c)^x，d'=e^d

Considering x=0, f(0)=1Knowable d'=1

Let e c=a again, you can get it.

f(x)=a^x

is an exponential function.

f'(x)=lna*a^x

Let x=0, it can be found'(0)

Substituting this gives us the exponential function as.

f(x)=e^[f'(0)*x]

Let x1 = 0, then we get f(0+x2)=f(0)·f(x2) so f(0)=1

and it is known that f(x) is a monotonic function on r.

So when x>0, f(x) > 1

Then let x1+x2=0, then f(x1)·f(x2)=1, i.e., f(-x)=1 f(x).

So f(x) Evergrande is at 0

It is also known that f(x) is a monotonic function on r.

So f(x) is an exponential function.

Therefore, the proposition is proven.

From the condition, knowing that f(0)=f(0)f(0), we get f(0)=0 or f(0)=1 If f(0)=0, then for any x, there is f(x)=f(x)f(0)=0 which is monotonically inconsistent with the function. Therefore, f(0)=1

and f'(x)=lim(h->0)[(f(x+h)-f(x))/h]lim(h->0)f(x)[(f(h)-f(0))/h]f(x)f'(0)

Solving this differential equation gives f(x)=ce [f'(0)x] and f(0)=1, so c=1, so f(x)=e [f'(0)x]

The proof process is as follows:

Roughly let me talk about the idea, for this type of proof, first of all, don't be too troublesome, compare two by two and then find the size, basically need 3 comparisons.

The basic idea of comparing the size of two functions, assuming the functions f(x), g(x), then establish a function h(x) = f(x)-g(x), find the derivative of h(x), and then calculate the increase and decrease of h(x) in the defined domain, and calculate the maximum minimum.

The derivable convex function has and only one maximum, so the minimum value is reached at the interval boundary.

If you still don't understand, you can consider g, which is obviously positive first and then negative, so g increases first and then decreases, and the minimum value is reached at the boundary, and the value inside the interval is the minimum value.

Similarly, a derivable concave function has and only one minima, which is the minimum of the concave function.

Because the hand-like plexus (x) (x)=f(x) g(x),f(x) g(x) increases monotonically in the area (a,b), the functions (x)=max and (x)=min also increase monotonically in the interval (a,b).

1, f1(x)=ax 2, substitute (1,1) to get a=1, so f1(x)=x 2, f2(x)=k x, and y=x are connected to the intersection point.

Root number k, root number k) or (-root number k, -root number k) Because the distance between the two intersection points is 8, the distance between the two points is obtained by the formula of the distance between the two points: 8k=8 under the root number, so 8k=64, so k=8 so f2(x)=8 x, f(x)=x 2+ 8 x,2, because f(x)=f(a) so x 2+ 8 x=a 2+ 8 a, the shift gets (x 2-a 2) + (8 x-8 a) = 0, (x+a)(x-a) + (8a-8x) (ax)=0

Mention the common factor to get (x-a) (x+a-8 ax) = 0, get x-a=0 or x+a-8 ax=0, x+a-8 ax=0 to ax 2+a 2x-8=0, because a>3

So its discriminant formula =a 4+32a>0, so x+a-8 ax=0 has two unequal real roots, and because x-a=0 gives the root x=a

So the equation f(x)=f(a) about x has three real solutions.

Proof: f(0)=f(0+0)=f(0)+f(0)=2f(0) f(0)=0

f(0)=f(x-x)=f(x)+f(-x)=0, i.e. f(-x)=-f(x).

The function y=f(x) is an odd function.

Take any x10 on the defined field r

When x 0 is again, f(x) 0 is constant.

f(x2-x1)<0

i.e. f(x2)-f(x1)<0

The f(x2) function y=f(x) is a subtraction function over r.

If f(1)=-670, find the maximum value of f(x) at [-3,3] from the previous proof, because f(x) is a subtraction function on r, so the maximum value is obtained at x=-3 and f(a+b)=f(a)+f(b).

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

Again, the function y=f(x) is an odd function.

f(-1)=-f(1)

f(-3)=-3f(1)=-3*(-670)=2010

Proof that f(a+b)=f(a)+f(b), let a=b=0, get f(0)=f(0)+f(0), that is, f(0)=0 let b=-a, get f(0)=f(a)+f(-a)=0, i.e., f(-x)=-f(x), and define the domain r

So, f(x) is an odd function.

2.Set x1>x2then x1-x2>0, so, f(x1-x2)<0 So, f(x1-x2)=f(x1)+f(-x2)=f(x1)-f(x2)<0

i.e. f(x1) So, f(x) is a subtraction function over r.

Let b = 0, f(a) = f(a) + f(0), so f(0) = 0.

Let b = -a, f(0) = f(a) + f(-a). So f(a)=-f(-a), so it is an odd function.

Let a=x, b=-x. So f(x)+f(-x)=f(0)=0. When x<0, -x>0, f(-x)<0, then f(x)>0

Let b=x,a>0, then f(x+a)-f(x)=f(a)<0, so it is a decreasing function.

The second step on the 1st floor is not done entirely discreetly.

Set x1>x2>0then x1-x2>0, so, f(x1-x2)<0 So, f(x1-x2)=f(x1)+f(-x2)=f(x1)-f(x2)<0

That is, f()-x2>0, f(-x1+x2)=f(-x1)+f(x2)=-f(x1)+f(x2)=-<0, so f(x1)-f(x2)>0, must be decreasing in (negative infinity, 0).

So decrementing on r.

Please enlarge the graph, obviously there is no 2y3=y1+y2 (it means that y3 is the ordinate of the midpoint of (x1,y1)(x2,y2)).