The known function f x e x 1 2 x 2 ax 1, where a is a real number

The first question is not difficult, there is no need to say more, some netizens have already given the answer before.

The second question is now discussed.

Let y1=f1(x)=e x, y2=f2(x)=x 2 2+ax+1, then f(x)=f1(x)-f2(x).

When x 1 2, f(x) 0 is constant, i.e., f1(x) f2(x) is constant.

As you can see from the graph, both functions pass through point (0, 1), and at the beginning, f2(x) increases a little faster, and later the exponential function f1(x) = e x increases a little faster, and the two have an intersection. To the right of the intersection, there must be f1(x)>f2(x).

The conclusions reached intuitively in the diagram cannot be used as strict proof, so the above conclusions need to be proved.

f1`(x)=e^x, f1`(0)=1;

f2`(x)=2x+a, f2`(0)=a；

When a>1, there is obviously a region near 0, f1 (x)f2 (x), but since f1(x) in the previous interval if a<1, then from 0 there is f1 (x)> f2 (x), and f1(0)>f2(0), so when x>0, there must be f1(x)>f2(x). (This conclusion is obvious, if you want to prove it strictly, the easiest way is to use the lagrange median value theorem, f(0)=f1(0)-f1(0)=1-1=0, when x>0, f1 (x)> f2 (x), so f(x)=f1 (x)-f2 (x)>0.) For any x>0, f(x)=f(x)-f(0)=x*f( 0, 0, x), the product f1(x)>f2(x).

Obviously, the larger the a, the greater the abscissa value of the intersection of f1(x) and f2(x) in (0, +). The critical case is that the abscissa of the intersection point is x=1 2, in this case, f1(1 2)=e (1 2)= e, and f1(1 2)=f2(1 2) obtains:

e=1 2*(1 2) 2+a*1 2+1, solution: a=2 e-9 4( >1).

To make x[1 2, +,f1(x) f2(x), just a2e-9.

Therefore, the range of values for a is (- 2 e-9 4).

1) a=-1 2.

f(x)=e^x-x^2/2+x/2-1

Define the domain r derivative f'(x)=e^x-x+1/2

f'(1)=e-1/2

f(1)=e-1/2+1/2+1=e-1

So the equation is y-e+1=(e-1 2)(x-1).

2e-1)x-2y-1=0

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

Continue to find the double guide f''(x)=e^x-1

According to f''(x) can know f'(x) decreases on (negative infinity, 0) and increases on [0, positive infinity).

So the minimum value f'(0)=1-a

Categorical discussions. When 1-a>=0, i.e., a=<1, f(x) increases monotonically, so according to the title, f(1 2)=e (1 2)-1 8-a 2-1>=0

The solution is a = <2 root number e-9 4

In summary, a=<1

When 1-a<0, a>1.

Let f'(x) The root is x1, x2(x1>x2) (this equation can't be solved, I'll express it with letters).

So f(x) increases on (negative infinity, x2), (x1, positive infinity) and decreases on [x1, x2].

So the minimum value f(x2) > 0

Solve another range.

This problem is a bit over-the-top in the end, and the transcendent equation will not be solved. Look at other people's ...

Participate in the dry and stupid test to defeat the stove.

f(x)=-a 2x 2+ax+c=-a 2(x-1 dress up 2a) 2+c+1 4a 1 2

f'(x)=e^x-a

a 0 hour f'(x) >0 f (x) when a>0 is monotonically incremented in the defined domain'(x)=0 then x=lnaxlna f'(x)>0 f(x) monotonically increasing to sum up.

a 0 f(x) is monotonically increasing within the defined domain.

a>0 f(x) monotonically decreasing in (-lna) and increasing monotonically in (lna, ).

Derivative of the function f(x) yields f'(x)=e x-a, then when a 0, f'(x) >0, then f(x) is monotonically increasing in the defined domain;

When a>0, let f'(x)=0, then x=lna, thus, xlna, f'(x) >0, f(x) monotonically increasing.

In summary, a 0, f(x) is monotonically increasing within the defined domain;

a>0 f(x) monotonically decreasing in (-lna) and increasing monotonically in (lna, ).

Is there any learning in seeking guidance? Direct derivation.

If you haven't learned, divide f(x) into g(x)=e x, h(x)=ax+1, and then do it, draw a picture to see the intersection, and think for yourself.

When a>0, 1+a>1,1-a<1,f(1+a)=-(1+a)-2a=-1-3a, f(1-a)=2(1-a)+a=2-a

f（1-a）=f（1+a）

2-a =-1-3a

a=-3 2 contradicts a>0.

When a<0, 1+a<1, 1-a>0

f(1+a)=2(1+a)+2a=2+4af(1-a)=-(1-a)-2a=-1-a 2+4a=-1-a,a=-3 5 is in line with the topic.

a=-3/5

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

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

Order f'(x)>0 i.e., e x-2>0, then the monotonic increase interval is x>ln2;

Order f'(x) <0 i.e. e x-2<0 then the monotonic subtraction is x0, then f'(x)=-2x＋e^x＋2a

So f'(x)=f(x)

(1) Solution: f(x)=ex-2x+2a,x r, f (x)=ex-2,x r

Let f (x) = 0 give x = ln2

So when x changes, f(x), f(x) changes as follows:

x(- ln2)ln2(ln2,+ f (x)-0+f(x) monotonically decreasing 2(1-ln2+a) monotonically increasing Therefore, the monotonically decreasing interval of f(x) is (- ln2), and the monotonically increasing interval is (ln2,+ f(x) at x=ln2, and the minimum is f(ln2)=eln2-2ln2+2a=2(1-ln2+a).

2) Proof : Let g(x)=ex-x2+2ax-1,x r, then g (x)=ex-2x+2a, x r know from (1) that when a ln2-1, the minimum value of g (x) is g (ln2)=2(1-ln2+a) 0 so for any x r, there is g (x) 0, so g(x) increases monotonically in r So when a ln2-1, there is g(x) g(x) g(0) for any x (0,+).

And g(0)=0, so that for any x (0,+ g(x) 0 is ex-x2+2ax-1 0, so exx2-2ax+1

Let a be a real number, and the function f(x)=e x-2x+2a,x r, the bounty score: 0 - 14 days and 22 hours until the end of the problem(1)Find the monotonic interval and extreme value of the function (2)Verify that when a ln2-1,x 0, e x x 2 2ax+1

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

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

Order f'(x)>0 i.e., e x-2>0 then the monotonic interval is x>ln2;

Order f'(x)<0 i.e., e x-2<0 then the monotonic interval is x0, then f'(x)=2x-e x-2=-(e x-2x+2a)+2a-2,=-f(x)+2a-2, obtained from (1), f(x)>=f(x)min =2-ln4+2asuoyi : f'(x)=-f(x)+2a-2 =< -2-ln4+2a)+2a-2=ln4-4<0,suoyi: f(x) monotonically subtracts in the defined domain.

When x=0, f(0)=0, suoyi: f(x) is monotonically reduced in the definition domain.

When x=0 f(0)=0, suoyi: f(x)x 2-2ax+1

Anyway, this answer won't be a standard answer.

When a ln2-1, f(x)=e x-2x+2a>e x-2x+2in2-2 makes h(x)=e x-2x+2in2-2,h'(x)=e x-2 in the decreasing interval of (0,ln2] less than or equal to 0(h(x)), [ln2,+infinity) is greater than or equal to 0(h(x)), h(x)min=h(in2)=0, so f(x)=e x-2x+2a>e x-2x+2in2-2=h(x)>=0 let f(x)=e x-x 2+2ax-1,f'(x)=f(x)>0, so f(x) is always an increasing function at x 0, so f(x)>f(0)=0, i.e., e x-x 2+2ax-1>0, i.e. e x x 2 2ax+1

Proof:

The constructor g(x) = (e x)-x +2ax-1 x r derivative, g'(x)=(e^x)-2x+2a.

g'(x)=f(x)

2] Function f(x) = (e x) - 2x + 2ax r derivative, f'(x)=(e^x)-2.

by f'(x)=(e x)-2=0 gives x=ln2 and when x ln2, e x e (ln2)=2x ln2, e x e (ln2)=2

On (-ln2), f'(x)＜0.At this point, f(x) decreases over (- ln2).

On (-ln2), f'(x) 0, at which point f(x) increments on [ln2,+.

When x=ln2, the function f(x) obtains the minimum value, and f(x)min=f(ln2)=2-2ln2+2a=2(a-ln2+1).

When a ln2-1, a-ln2+1 0

i.e. when a ln2-1, f(x)min 0

Or rather, when a ln2-1, g'(x) 0 At this point, the function g(x) is incremented on r.

When x 0, there is always g(x) g(0) (easy to know g(0)=0) that is, there is constant (e x)-x +2ax-1 0

When a ln2-1 and x 0, there is always :

e^x＞x²-2ax+1

f(x)=e^x(x^2-ax+a)

f'(x)=e^x(x^2-ax+a)+(2x-a)e^x=e^x[x²-(a-2)x]=xe^x(x-a+2)

When f'(x) > 0, f(x) increases monotonically.

i.e. xe x(x-a+2)>0

Because e x > 0 early.

x(x-a+2)>0

and staring at it because a>2

Get x>a-2 or Kailu touching x

(1) When a=0:

f(x)=x｜x｜

f(-x)=(-x) -x =-x x =-f(x) is an odd function at this time. When a ≠

At 0: f(-x)=-x -x-a =-x xa is not necessarily related to f(x).

In this case, it is a non-odd and non-even function.

2) When a=2

f(x)=x｜x-2｜

f(8)=8*6=48

f-1(8)=1/48

..F-1(8) is it the reciprocal of it, or the value of its inverse function, I'm looking for the reciprocal value of it.