If the function y = e (A-1) x + 4x (x ∈ R) has an extreme point greater than zero, the range of real number a is () A. a＞-3 B. a＜-3 C. a＞−1 three D. a＜−1 three

If the function y = e (A-1) x + 4x (x ∈ R) has an extreme point greater than zero, the range of real number a is () A. a＞-3 B. a＜-3 C. a＞−1 three D. a＜−1 three

Because the function y = e (A-1) x + 4x,
So y '= (A-1) e (A-1) x + 4 (a < 1),
So the zero of the function is x0 = 1
a−1ln4
1−a，
Because the function y = e (A-1) x + 4x (x ∈ R) has extreme points greater than zero,
So x0 = 1
a−1ln4
1 − a > 0, i.e. ln4
1−a＜0，
The solution is: a < - 3
Therefore, B

If the function FX = ax ^ 3-bx + 4, when x = 2, the function FX has the extreme value - 4 / 3 1. Find the analytical formula of the function If the function FX = ax ^ 3-bx + 4, when x = 2, the function FX has the extreme value - 4 / 3 1. Find the analytical formula of the function 2. If the equation f (x) = k has three different roots, find the real number K Range of values

F (2) = 8a-2b + 4 = - 4 / 3, if there is an extreme value, f '(2) = 12a-b = 0, so a = 1 / 3, B = 4f (x) = 1 / 3x ^ 3-4x + 4, if f' (x) = x ^ 2-4 = 0, x = 2 or x = - 2, f (- 2) and f (x) are two extreme values, f (- 2) = 8 + 4-8 / 3 = 26 / 3, so K has three different roots between intervals (- 4 / 3, 26 / 3)

If the definition field of the square of ax minus ax plus 1 / A under the function y = root is r, what is the value range of the real number a?

∵Y=√（ax ²- Ax + 1 / a) the definition field is r
∴ax ²- Ax + 1 / a ≥ 0
∴a＞0, Δ= a ²- 4≥0
∴a≥2

Function y = root (AX) ²+ If the definition field of 3ax + 1) is r, the value range of real number a is To the process, there are extra points

Answer:
The definition field of y = √ (AX ^ 2 + 3ax + 1) is the real number range R
Therefore: ax ^ 2 + 3ax + 1 > = 0 is always true
1) When a = 0, 0 + 0 + 1 > = 0 is constant, which is in line with the meaning of the question
2) When a < 0, the opening of parabola g (x) = ax ^ 2 + 3ax + 1 is downward, and there is always x, so that G (x) < 0, which is inconsistent with
3) When a > 0, the parabola g (x) = ax ^ 2 + 3ax + 1 > = 0, the opening is upward and always above the x-axis
There is at most one intersection with the X axis, so:
Discriminant = (3a) ^ 2-4a < = 0,9a ^ 2-4a < = 0,0 < = a < = 4 / 9
To sum up, 0 < = a < = 4 / 9

Find the area of the plane figure enclosed by the parabola y ^ 2 = 2x and the straight line y = 4-x? Do with calculus Just write the answers

1. First find the intersection of parabola and straight line y ^ 2 = 2x y = 4-x (4-x) ^ 2 = 2x x ^ 2-10x + 16 = 0x1 = 2 Y1 = 4-2 = 2 points (2,2) x2 = 8 y2 = 4-8 = - 4 points (8, - 4) 2. Then find the integral y. The range of integral y is from - 4 to 2 (upper 2, lower - 4, the same below) y ^ 2 = 2x x = y ^ 2 / 2Y = 4-x x = 4-y ∫ (- 4,2) (4-y ^ 2 / 2) dy = (4y-1 / 2Y

Parabolic equation expression

Parabola equation refers to the trajectory equation of parabola, which is a method of expressing parabola with equation. Parabola can be drawn according to the equation of parabola on the geometric plane. The specific expression of the equation is y = a * x * x + b * x + C (1) a ≠ 0 (2) a > 0, then the opening of parabola faces upward; A < 0, then the parabolic opening faces downward; (3) extreme point: (- B / 2a, (4ac-b * b) / 4A); ⑷ Δ= b*b-4ac, Δ＞ 0, the image intersects the x-axis at two points: ([-b- √ Δ]/ 2a, 0) and ([- B + √ Δ]/ 2a,0）； Δ＝ 0, the image intersects with the x-axis at one point: (- B / 2a, 0); Δ＜ 0, the image has no intersection with the x-axis; if the parabola intersects the y-axis as the positive half axis, then c > 0. If the parabola intersects the y-axis as the negative half axis, then C