## Let function f (x) = (square of X + ax + b) e to the power of X (x ∈ R). If a = 2 and B = - 2, find the extreme value of function f (x)

f(x)=(x^2+2x-2)e^x; Derivation: F '(x) = (2x + 2) e ^ x + (x ^ 2 + 2x-2) e ^ x = (x ^ 2 + 4x) e ^ X; Let f '(x) = 0; The solution is x = 0 or - 4; Similarly, f '' (x) = (2x + 4) e ^ x + (x ^ 2 + 4x) e ^ x = (x ^ 2 + 6x + 4) e ^ X; f''(0)=4>0; f''(-4)=-4*e^(-4)

### The image passing point (π) of the known function f (x) = asinxcos x-2cos 2x + 1 8，0)． (I) find the value of real number a; (II) if x ∈ [0, π) and f (x) = 1, find the value of X

(I) f (x) = asinxcosx − 2cos2x + 1 = a2sin2x − cos2x (3 points) according to the meaning of the question, f (π 8) = 0, that is, a2sin π 4 − cos π 4 = 0, the solution is a = 2 (6 points) (II) according to (I), f (x) = sin2x − cos2x = 2Sin (2x − π 4). According to the meaning of the question, sin (2x − π 4) = 22 (9 points) because 0 ≤ x

### If the function f (x) = the 4th power of AX + the square of BX + C satisfies f (- 3) = 7, what does f (3) = mean?

F (x) = ax ^ 4 + BX ^ 2 + C is an even function

f(-x)=f(x)

So f (3) = f (- 3) = 7

### It is known that the function f (x) = the third power of X - the square of 1 / 2 * x + BX + C. (1) if f (x) is an increasing function on (- ∞, + ∞) The function f (x) = the third power of X - the square of 1 / 2 * x + BX + C. (1) if f (x) is an increasing function on (- ∞, + ∞), find the value range of B. (2) if f (x) obtains the extreme value at x = 1 and X ∈ [- 1,2], the square of F (x) < C is constant, and find the value range of C Thank you again

f'（x）=3x^2-x+b>=0 Δ= 1/12

2）f'（1）=0 b=-2

When x ∈ [- 1,2], the square constant of F (x) < C holds

Case by case discussion

### It is known that the vertex coordinates of the quadratic function y = ax * 2 + BX + C (a ≠ 0) are (- 1, - 3.2) and some images are shown in the figure. From the image, we can know the univariate quadratic power of X

== I'm doing this problem, too. Oh, my God. Isn't it TVT. OK, try again

Firstly, from the vertex formula: - B / 2A = - 1, we know that: B = 2A

4ac-b^2/4a=3.2

∵b=2a

∴4ac-4a^/4a=3.2

4a(c-a)/4a=3.2

c-a=3.2

c=3.2+a

—————————————————————————

Then y = ax ^ 2 + 2A * x + 3.2 + A

Bring in y = 0, X1 = 1.3

Get: 1.69a + 2.6a + A + 3.2 = 0

== then LZ can think about it. Or let's discuss it. I'm stuck here

————————————————————————

== forget it, I'll figure it out. I'll round you up if there's a decimal anyway

5.29a=-1.2

a=-0.6

Get: - 0.6x ^ 2-1.2x + 3.2-0.6 = 0

x^2+2x-4.3=0

(x^2+2x+1-1)=4.3

(x+1)^2=5.3

x1+1=2.3

X1 = 1.3 (rounded)

x2+1=-2.3

x2=-3.3

∴x2=-3.3

Hey, we finally finished this problem = = you have to figure it out for yourself, LZ. You can figure it out by yourself

If later people look at this question again, I think it should be almost the same. It's just that it's worthy of that picture~

——————————————————————————————————————

It turns out to be right = v=

The teacher ticked

### Quadratic function y = the quadratic power of AX + BX + C. If A-B + C = 0, what point does its function have to go through?

A-B + C = 0 means that when x = - 1, y = A-B + C = 0

So it must pass (- 1,0)