Let the power function f (x) = k times the image of x ^ a passing through the point (1 / 2, root 2 / 2), then K + a =?

Let the power function f (x) = k times the image of x ^ a passing through the point (1 / 2, root 2 / 2), then K + a =?

K = 1, a = 1 / 2, so K + a = 3 / 2
Because the function is a power function, k = 1
And because of passing point (1 / 2, radical 2 / 2), so a = 1 / 2
So K + a = 3 / 2
1 * (1 / 2) ^ (1 / 2) = radical 2 / 2
a=1/2, k=1
a+k=3/2
If the image of power function y (x) passes through points (9,3), then the value of F (25) is
Then f (x) = x ^ a
(9,3)
Then f (9) = 3
9^a=3=9^(1/2)
a=1/2
So f (25) = 25 ^ (1 / 2) = 5
Let the power function be y = x ^ a
Because it's over (9,3)
Then 3 = 9 ^ a = (3 ^ 2) ^ a = 3 ^ (2a)
Then 2A = 1
a=1/2
Then f (25) = (25) ^ (1 / 2) = 5 ^ [2 * (1 / 2)] = 5
Given the quadratic function y = x & # 178; - (M + 1) x + 9, if the value range of this function is [0, + ∞], find the value of real number m (with process)
The derivative of X is 2x-m-1 = 0,
From the range, we can get 0 = x & # 178; - (M + 1) x + 9
By solving the equations, we can get x = plus or minus 3, which are substituted into the judgment,
M = 5 for positive 3, M = - 7 for negative 3
The opening of function image is upward, and the value range is [0, + ∞),
That is to say, there is only one intersection point between the function and the X axis, so the discriminant of the equation x & # 178; - (M + 1) x + 9 is equal to 0,
That is △ = [- (M + 1)] ² - 4 × 1 × 9 = 0, M = 5 or M = - 7
Find the value of the power of 2012 of negative zero point 25 multiplied by the power of 2011 of 4
It's - 1 / 4
The problem of trigonometric function of acute angle
1. The size relationship of sin70 °, cos70 ° and tan70 ° is () write the process
2. In the triangle ABC, D is the midpoint of AB, DC is perpendicular to AC, and tanbac = 1 / 3, find the value of sina, cosa, Tana? Write the process specifically
Good answers are given
1) In the triangle ABC, D is the midpoint of AB, DC is perpendicular to AC, and tanbac = 1 / 3 indicates that the triangle ADC is a right triangle
1. Draw a picture
1、tan>sin>cos。 Firstly, Tan 70 is greater than 1. Secondly, when Tan 70 is greater than 45 and less than 90, sin > cos.
Second, let me draw a picture to study it
1. Tan 70 ° > sin 70 ° > cos 70 ° drawing method, very intuitive
Two
Tan > sin > cos
Compare sin and COS first, because cos70 = sin20 is less than sin70
Look at Tan again, tan70 = sin70 / cos70 > sin70 (cos70 is less than 1)
So tan is greater than sin and COS
The second question is wrong. Please check it and send it back to me
The quadratic function f (x) satisfies f (x + 1) = 2F (x) - X & # 178;
1 find the analytic expression of F (x). 2 if f (x) ≥ m is constant, find the value range of real number M
1,
Let f (x) = ax & # 178; + BX + C
Then f (x + 1) = a (x + 1) &# 178; + B (x + 1) + C
a(x+1)² +b(x+1)+c=2ax² +2bx+2c-x²
ax²+2ax+a+bx+b+c=(2a-1)x² +2bx+2c
ax²+(2a+b)x+a+b+c=(2a-1)x² +2bx+2c
So a = 2a-1, the solution is a = 1
2a+b=2b
B=2
a+b+c=2c
C=3
So the quadratic function f (x) = x & # 178; + 2x + 3
Two
If f (x) ≥ m is constant, i.e
x² +2x+3≥m
(x+1)²≥m-2
m-2≤0
m≤2
First simplify, then evaluate: [(2x + y) 2 - (2x-y) (2x + y)] / (2Y), where x = 2, y = - 1
[(2x + y) 2 - (2x-y) (2x + y)] / (2Y), = [4x2 + 4xy + y2-4x2 + Y2] / (2Y), = (4xy + 2Y2) / (2Y), = 2x + y, when x = 2, y = - 1, the original formula = 2 × 2 + (- 1) = 3
Two questions about acute angle trigonometric function
1. Cosa = 1:4, calculate angle a degree
2. Construct a right triangle with an angle of 15 degrees and find three trigonometric functions of 15 degrees
One
A=arccos(1/4)
Two
Triangle ABC, a = RT, C = 30 °, CD bisects C, intersects AB to D, ACD = 15 °
Suppose AB = 1, BC = 2, AC = √ 3
AD=X,BD=1-X
BC/AC=DB/AD
2/√3=（1-X）/X
X=2√3-3
CD=√（AC^2+AD^2)=3√2-√6
sin15°=AD/CD
=(2√3-3)/(3√2-√6)
=(√6-√2)/4
sin15°=(√6-√2)/4
cos15°=(√6+√2)/4
tan15°=2-√3
cot15°=2+√3
Let f (x) = x & # 178; - ax + 1
(1) If f (x) is not monotone in the interval [2,3], find the value range of real number a
(2) If the minimum value of F (x) in the interval [2,3] is 1, find the value of real number a
(3) If α and β are two of the equations f (x) = 0, find the minimum value of (α - 1) &# 178; + (β - 1) &# 178
1. If [2,3] is not monotone, it means that the vertex of the curve is in the interval [2,3], and the abscissa formula of the vertex is - B / 2a, that is, a / 2
Two
One
2 a=2
When a = 1, the minimum value is - 1
2 when x = 0, f (x) = 1, f (x) = x & # 178; - ax + 1 is upward, and the minimum value is 1, indicating that [2,3] is on the right side of the curve,
When x = 2, f (2) = 4-2a + 1 = 1, a = 2
3A & # 178; - AA + 1 = 0... Expansion
One
2 a=2
When a = 1, the minimum value is - 1
2 when x = 0, f (x) = 1, f (x) = x & # 178; - ax + 1 is upward, and the minimum value is 1, indicating that [2,3] is on the right side of the curve,
When x = 2, f (2) = 4-2a + 1 = 1, a = 2
3A & # 178; - AA + 1 = 0, B & # 178; - AB + 1 = 0 (α - 1) & # 178; + (β - 1) & # 178; = a (A-2) = (α - 1) & # 178; - 1 when a = 1, the minimum value is - 1
Simplify the evaluation: the square of (y + 2x) (2x-y) - (2x + y) + the square of 2Y, where x = 1, y = - 2
The square of (y + 2x) (2x-y) - (2x + y) + the square of 2Y
=(2x+y)(2x-y-2x-y)+2y^2
=-2y(2x+y)+2y^2
=-4xy
X = 1, y = - 2
-4xy=8