Let f (x) = ex (x2 + ax-a), where a is a constant. (I) when a = 1, find the tangent equation of F (x) at point (1, f (1)); (II) find the minimum value of F (x) in the interval [0, + ∞)

Let f (x) = ex (x2 + ax-a), where a is a constant. (I) when a = 1, find the tangent equation of F (x) at point (1, f (1)); (II) find the minimum value of F (x) in the interval [0, + ∞)

(I) from F (x) = ex (x2 + ax-a), f ′ (x) = ex [x2 + (a + 2) x] When (2) f, e = 1 So the tangent equation of curve y = f (x) at point (1, f (1)) is y-e = 4E (x-1), that is y = 4ex-3e (6 points) (II) Let f '(x) = ex [x2 + (a + 2) x] = 0, the solution is x = - (a + 2) or x = 0 (8 points) when - (a + 2) ≤ 0, i.e. a ≥ - 2, on the interval [0, + ∞), f '(x) ≥ 0, so f (x) is an increasing function on [0, + ∞). So the minimum value of F (x) is f (0) = - A; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp (10 points) when - (a + 2) > 0, i.e. a < - 2, the changes of F '(x), f (x) with X are as follows: x 0 (0), (a + 2)) - (a + 2) (- (a + 2), + ∞) f' (x) 0 - 0 + F (x) f (0) ↘ f（-（a+2）） A kind of It can be seen from the above table that the minimum value of function f (x) is f (- (a + 2)) = a + 4ea + 2 (13 points)
Find the range of y = 6 (cosx) ^ 4 + 5 (SiNx) ^ 2-4 / cos2x
cos²x=(1+cos2x)/2
So (cosx) ^ 4 = (cos2x + 1) & sup2 / / 4
sin²x=(1-cos2x)/2
So molecule = 6 (cos2x + 1) & sup2 / 4 + 5 (1-cos2x) / 2-4
=(3cos²2x-cos2x)/2
So y = (3cos2x-1) / 2
Where the denominator cos2x is not equal to 0
So cos2x = 1, ymax = 1
Cos2x = - 1, ymain = - 2
Cos2x is not equal to 0, y is not equal to - 1 / 2
So the range [- 2, - 1 / 2) ∪ (- 1 / 2,1]
cos2x≠0,x≠k∏±∏/4
y=6(cosx)^4+5(sinx)^2-4/cos2x
=[6(cos2x+1)^2/2+5(1-cos2x)/2-4]/cos2x
=[3(cos2x+1)^2-3-5cos2x)]/(2cos2x)
=[9cos^2(2x)+cos2x]/(2cos2x)
=1/2(9cos2x+1)
ymax=5,ymin=-4
cos2x≠0,y≠1/2
Range - 4
The point represented by a number a on the number axis is a. when point a moves three unit lengths to the left on the number axis, it is point B and ab are opposite to each other! What is a?
One point five
One point five
Would you be my cup of tea
"Be one's cup of tea" is used to mean "something that someone is interested in or likes. It is often used to imply a lover.
Would you like to be my lover?
Popular point:
Will you be my dish?
Are you my cup of tea?
Given the function f (x) = x / ax + B (A.B is a constant and ab ≠ 0)
If f (2) = 1, f (x) = x has a unique solution, then the analytic expression of y = f (x) is
——y=2x/x+2(x≠2)
F (x) = x has a unique solution What does this sentence mean? What are the hidden conditions?
f(2)=2/（2a+b）=1 a=(2-b)/2
f(x)=x/(ax+b)=x ax^2+(b-1)x=0
Because there is a solution
△=（b-1）^2-4a*0=0
(b-1)^2=0
B=1
a=(2-1)/2=1/2
f(x)=x/(x/2+1)=2x/(x+2)
There is only one solution, which is delta = 0~
The definition and range of cos2x / (SiNx cosx)
Domain of definition: SiNx cosx ≠ 0, X ≠ K π + π / 4
Range: [- √ 2, √ 2], because:
cos2x/（sinx-cosx）
=（cos²x-sin²x） /（sinx-cosx）
=-(sinx+cosx)
=-√2sin(x+π/4)
On the number axis, if a point represented by a number is shifted to the right by 6 units of length and its opposite number is obtained, then the number is zero______ .
Let this number be X. according to the meaning of the question, x + 6 = - x, the solution is: x = - 3, so the answer is - 3
It's really your cup of tea?
cup of tea
A favorite person or thing
This is exactly what you want
It's really your style
It's really to your taste
Given f (x) = (x + 1) ^ 2 + SiNx / x ^ 2 + 1, if a = f (LG2) B = f (LG1 / 2), then a.a-b = 0 B.A + B = 2 c.a-b = 1 D.A + B = 1
The molecule is (SiNx + 1)
Obviously, if LG1 / 2 = - LG2, then a = f (LG2) = [(LG2 + 1) ^ 2 + sin (LG2)] / (LG & # 178; 2 + 1) and B = f (LG1 / 2) = [(- LG2 + 1) ^ 2 + sin (- LG2)] / (LG & # 178; 2 + 1), then a + B = [(LG2 + 1) & # 178; + sin (LG2) + (- LG2 + 1) & # 178; + sin (- LG2)] / (LG & # 178; 2 + 1) and sin (LG) = [(LG2 + 1) & # 178; 2 + 1)] / (LG & # 178; 2 + 1)
Such a simple question. I don't know how to answer.
F (x) = (SiNx + cosx)'2 + cos2x. Find the minimum period and range of F (x)
F(X)=(sinx+cosx)^2+cos2x=1+sin2x+cos2x=(√2)sin(2x+π/4)+1
Therefore, the minimum positive period is 2 π / 2 = π;
Since - 1 ≤ sin (2x + π / 4) ≤ 1, 1 - (√ 2) ≤ f (x) ≤ 1 + (√ 2),
So, the range is [1 - (√ 2), 1 + (√ 2)]