Let the function f x e x 1 x ax 2 if x 0 and f x 0 and find the range of values of a

Knowable. f(0)=0

f'(x)=e^x-1-2ax

f'(0)=0

f''(x)=e^x-2a

f''(0)=1-2a

When a<=1 2.

For any x>=0

f''(x)=e^x-2a>=1-1=0

So f'(x) is an increment function within the defined domain.

f'(x)>=f'(0)=0

So f(x) is the increasing function, f(x)>=f(0)=0 is the rigor of the proof, and the following proves that x exists when a>1 2, such that f(x) is less than 0 when a>1 2.

The original proposition does not hold when 01 2 exists.

So a<=1 2

f(x)=x(e^x-1)-ax2

So. f’(x)=

e^x(x1)-2ax-1

and f(0)=0

To make. f(x)>= is constant on x>=0.

Rule. f'(x)>=0 is to be constant.

Namely. e^x(x

1)-2ax-1>=0

Here I don't think it is possible to separate a: a<=

e x(x1)-1) (2x), let t(x)=e x(x1)-1) (2x), then t'(x)=e x*x 2

e^x*x-

e x-1, so that t'(x)=0, gives x=0, and t(x) in x cannot be 0) so g(x)=

e x(x1)-2ax-1, i.e. g(x)>=0 and g(0)=0, so g'(x)>=0 must be constant.

g’(x)=

e^x*xe^x-2a>=0

At this point, you can separate a).

So a<=e x(x

Let p(x)=e x(x

then p'(x)=(e x*x

e x) 2, let p'(x)=0

Obtained x=-1, we can see that x=-1 is the p(x) minimum.

And x>=0, then the minimum value of p(x) is p(0)=1 2, so a<=1 2

Dude doesn't know if it's right. This problem is computationally intensive, and it is necessary to find a second derivative. I saved some steps. If you have questions, you can communicate.

Solution: When x>=0, f(x)>=0

That is, the value of a satisfies the minimum value f(x)=0

f(x)=e^x-1-x-ax^2=0

f'(x)=e^x-1-2ax=0

f"(x)=e^x-2a>=0

f(x)-f'(x)=ax 2-(2a-1)x=0x=2a-1 or. x=0

f(0)=0)).

2a-1=x>=0

a>=1/2

The maximum value of note a is at least 1 2

f"(x)=e^x-2a>=0

When x>=0 is all true, so.

e^0-2a>=0

a<=1/2

Note that the maximum value of a can only be 1 2

The value range of a is a<=1 2

Equivalent to "e x-1-x-ax 2 0."

About. x 0 is constant", solve a from the inequality, and then find the minimum value of the right function, a(e x-1-x) x, x 0, and thus a

e^x-1-x)/x²]min

Let g(x) = (e x-1-x) x, x 0, find the minimum value.

If you launch this result assuming f(x)=0, what about f(x)=other values? This is where your problem lies, you can't push it big from a small one!

Weak explosion. I'm not done.

The difficulty is when it is >.

Derivative f(x) yields f'Songyun(x)=e x-1-2ax so when a0 is constant, so the shed beam f(x) is an increasing function, then the minimum value of f(x) is f(0), f(0)>=0, obviously f(0)=0, so a0 you can draw e x-1=2ax first, first draw the left side and the right side of the base into 2 images, recorded as y1=e x-1, y2=2ax, here I won't.

f(x)=x(e x-1)-ax2 so f' disadvantage (x) = e x(x+1)-2ax-1 and f(0)=0 to make f(x)>= constant on the rental sedan x>=0 then f'(x)>=0 to be constant immediately e x(x+1)-2ax-1> let chaos = 0 (here I don't think a can be separated: a=0 and g(0)=0, so g'(x)>=0 should always be established g'(x)= e x*x+ e x-2....

Deriving f(x) yields the empty orange prying f'(x)=e x-1-2ax so when a0 is constant, so f(x) is an increasing function, then f(x) minimum is f(0), f(0)>=0, obviously f(0)=0, so a0 you can draw e x-1=2ax first, first draw the left and right sides separately into 2 images, marked as y1=e x-1, y2=2ax, here I won't make the image, you draw it yourself. I'll explain the point in detail where the slag looks at the slope of y2, since the y1 image is always on top of y2 then obviously f'(x)>=0 is constant, there is no doubt about this, then since y1 is 1 when x=0, then, 2a1, then y1, y2 have 2 intersections. One is 0, and the other can't be asked, (hehe, you teachers can't ask for it either,) but it's okay.

You look at the image, and now Dou Jing has two intersections, one is 0 and the other is x1, so on [0, x1] y2 is higher than y1, so f'(x) is less than 0, the function is monotonically decreasing, and then you are looking at f(0)=0 is constant, so this means that on [0,x1] f(x),6,a>=1 2

It can be found using the derivative. ,0,

The constructor g(x)=f(x)-ax, x 0 because: if pure x=0, g(0)=1-1-0=0 and f(x) ax for all x 0s, then g(x) must be incremented at 0, i.e., g (x) is positive at x=0 and g (x)=e x+e (-x)-a 2 e x*e (-x) -a=2-a, and so on if and only if x=0. So g .

If f(x)>=0 when x>=0, find the range of a(x)=x*(e x-1)-ax 2

So f'(x)=e^x-1+x*e^x-2ax=(x+1)e^x-2ax-1

Then when x=0, there is: f'(x)=0。and f(0)=0 is known to be f(x) 0 when x 0

Therefore, it must be satisfied at x 0, f'(x) 0 [because this is the only way to ensure that f(x) is incremented at x 0 and f(x) f(0)=0].

Then: f''(x)=e x+(x+1)e x-2a=(x+2)e x-2a is greater than or equal to zero at x 0.

So, (0+2)*e 0-2a 0

then, a 1

Solution: f(x)=x(e x-1)-ax ==> f(0) = 0

If f(x) is an increment function on (0, +, i.e. f'(x)>0, then for any x>0, there is:

f(x) >f(0) ==>f(x) >0

Thus making f(x) 0 on the closed interval [0, +

f'(x) = (x+1)e^x -1 - 2ax ==> f'(0) = 0

In the same way, if in (0, +f''(x) >0, then f'(x) 0 on [0, +] is guaranteed

f''(x) = xe^x +2e^x- 2a

Order f''(x) >0 is held at (0, +, then 2a 2e 0< xe x +2e x ==> a 1

When a 1, f(x) is an increment function over (0, +, thus x 0.

f(x) = x(e^x-1) -ax² ≥0

Conclusion: The value range of a is a 1

Idea: Separate parameters.

When x=0, f(0)=0, f(x) 0 is true;

When x>0, f(x)0 can be turned to.

a≤(e^x-1)/x

Let g(x)=(e x-1) x, then a [g(x)]min,x>0

And g'(x)=[xe^x-(e^x-1)]/x²=[(x-1)e^x+1]/x²

Let h(x)=(x-1)e x+1,x 0

h'(x)=e x+(x-1)e x=xe x 0 so h(x) in [0, ) is the increasing function, h(x) h(0)=0 and thus g'(x)=h(x) x 0, g(x) at (0, ) is an increment function.

So a lim(x 0)g(x)=0

i.e. a 0

Solution: f(x)=e x-1-x-ax 2 ==> f(0) = e 0 -1-0 -a*0 = 0

If f(x) is an increment function on (0, +), then for any x>0, there is:

f(x) >f(0) ==>f(x) >0

Thus making f(x) 0 on [0, +

f'(x) = e^x -1 - 2ax

Same f'(0) =0；If in (0, +f''(x) >0, then f'(x) >0

f''(x) = e^x - 2a

Order f''(x) >0, then 2a e 0< e x ==> a 1 2

So when a 1 2, f(x) is an increasing function on (0, +, and thus x 0.

f(x) = e^x-1-x-ax^2 ≥ 0

It is easy to know that f(0) = 0

f'(x)=e^x-1-2ax

f'(0)=0

f''(x)=e^x-2a

f''(0)=1-2a

When a<=1 2.

For any x>=0

f''(x)=e^x-2a>=1-1=0

So f'(x) is an increment function within the defined domain.

f'(x)>=f'(0)=0

So f(x) is the increasing function, f(x)>=f(0)=0 is the rigor of the proof, and the following proves that x exists when a>1 2, such that f(x) is less than 0 when a>1 2.

The original proposition does not hold when 01 2 exists.

So a<=1 2

When x>=0, e x>=1.f(x)>=0, then ax 2+x+1<=e x.

When x=0, f(x)=0....

Then x>=0, when f(x)>=0, that is, the minimum value of f(x)" = 0.

f'(x)=e^x-1-2ax..

Discuss if a<=0, then x>=0, f'(x) >=0 satisfies the condition.

If a>0, then f when x=0'(x) <0, then f(0)=0, then when x > 0, f(x) will always be less than 0 when x is a little greater than 0. It is impossible to meet the conditions.

I hope you take a good look and understand. So in short, a<=0.

The taylor formula of e x is 1+ x n (n!n from 1 to +) is written in the form with peano remainder.

e x=1+x+x 2 2+o(x 2) o(x 2) denotes the higher-order infinitesimal of x 2.

From this we can see a 1 2 o'clock.

f(x)=e^x-1-x-ax^2

1-a)x^2+o(x^2)≥0

a>1 2.

f(x) is less than 0 when x tends to 0+

Therefore a 1 2

It is equivalent to "e x-1-x-ax 2 0 is constant for x 0", solve a from the inequality, and then find the minimum value of the right function, a (e x-1-x) x , x 0, so that a [e x-1-x) x ]min

Let g(x) = (e x-1-x) x, x 0, find the minimum value.

You'd better get the format out of the topic, so you won't be able to see what the topic means.

It's weak, it's not done, it's hard when it's >.