Math problem 10 on derivatives

Solution: Because 2f(x)+xf(x) x 2 ......These are discussed below:

1) When x= 0, substitute : f(0) 0

2) x 0, both sides are multiplied by x: 2xf(x)+x 2f(x) x 3, i.e.

x 2f(x)] x 3 0, so the function y= x 2f(x) is an increasing function on r+, and x 0, therefore: x 2f(x) 0 2f(0) = 0 , so f(x) 0

3) x 0, multiply x by x on both sides: 2xf(x)+x 2f(x) x 3, i.e.

x 2f(x)] x 3 0, so the function y= x 2f(x) is an increasing function on r-, and x 0, so x2f(x) 0 2f(0) = 0 , so there is also f(x) 0

To sum up, when x r, there is always f(x) 0 so choose a

— But ——— is ———

Multiple choice questions should do just that! By f(0) 0 i.e. excluding options b and c, it is clear that f(x)=x 2 +a(a 0) when the known condition 2f(x)+xf(x) x 2 holds, but.

f(x) x may not be true, so d is also wrong, so choose a

I don't know how to push.

Suppose f(x)=x +1 4

2f（x）+xf'（x）=2x²+1/2+2x²=4x²+1/2＞x²

then f(x) 0 is clearly not true; f(x) x is manifestly untenable; f(x) x is not true when x=-1 2;

So f(x) 0

Let the function f(x)=(a 2)x +[x+1) e x]-1(1) If a=0, find the monotonic interval of f(x). (2) If x 0 and f(x) 0 are constant, find the value range of a.

Solution: (1) When a=0, f(x)=(x+1) e x-1

Let f (x)=[e x-(x+1)e x] e (2x)=-x(e x) e (2x)=-x e x=0

The station point x=0

When x<0, f(x)>0; When x<0, f(x)>0;

Therefore, in the interval (- 0), f(x) increases monotonically; Monotonically decrease in the interval (0,+. x=0 is the maximum.

Maximum f(0)=0

2) At x 0, for the inequality f(x)=(a 2)x +[x+1) e x]-1 0 to be constant, it is necessary to make .

f (x)=ax-x e x=[(ae x-1) e x]x 0 is constant, that is, to make ae x-1 0, a 1 e x constant, x 0, 1 e x 1, so the value range of a is: a 1

f(x)=a 2 x 2+(x+1) e x-1 What is the problem?

Is f(x)=(a 2)*x 2+(x+1) (e (x-1))?

Or f(x)=(a 2)*x 2+(x+1) (e x)-1)?

The derivative of g(x) h(x) = -1 (x 2) + a x + 2x

Note that when x>=1, -1 (x 2)+a x+2x>=0, that is, a>=1 x-2x2 is constant, that is, the maximum value of 1 x-2x 2 is found, and because 1 x-2x 2 is a subtraction function, that is, when x=1, the maximum value is -1, that is, a>=-1

Derivative of g(x) = -1 (x 2) + a x + 2x

Divide and then discuss by category. Above.

The rate of change is the derivative, and the average rate of change should be the derivative f'(x) The quotient of the definite integral between [3,a] and (a-3).

Derivative function f'The integral of (x) is the original function f(x), so this definite integral is equal to f(a)-f(3), which is to find a when the median value of Glaranger is equal to 5.

That is, [f(a)-f(3)] (a-3)=(a 2-2a-3) (a-3)=5, the solution is: a=4 or a=3 (rounded), so the answer is (4).This answer fits the topic well and shows that my speculation is correct.

f(x)=a*b=x^2(1-x)+(x+1)t=x^2-x^3+xt+t

f'(x)=-3x^2+2x+t

f(x)=a*b is an increasing function over the interval (-1,1), i.e., f'(x) is greater than 0. on (-1,1).

f'The axis of symmetry of (x)=-3x 2+2x+t is: x=-2 (-6)=1 3, to get f'(x) is greater than 0 on (-1,1), then there is x1<-1,x2>1

x1=[-2+root(4+12t)] (-6)<-1x2=[-2-root(4+12t)] (-6)>1 solves: t>5

A function is an increasing or decreasing function in an interval, and the range of unknowns can be found using derivatives.

I've done the math and you're having a bit of a problem finding the vector a vector b. (x^2,x+1)

1-x,t)=-x 3+x 2+tx+t function f(x)=a*

b is an increasing function over the interval (-1,1), then there is f'(x)=-3x 2+2x+t 0 is constant on (-1,1). (Note to bring an equal sign, otherwise a case will be missed).

So there is t 3 x 2-2x = 3 (x-1 3) 2-1 3 so t 3 x 2-2x maximum.

And the maximum value of 3x 2-2x is less than 5

So t 5