Given that the function f (x) = x 2 + 2aX + 2, X belongs to [- 5,5], when a = - 1, find the maximum and minimum value of the function

Given that the function f (x) = x 2 + 2aX + 2, X belongs to [- 5,5], when a = - 1, find the maximum and minimum value of the function

When a = - 1
f(x)=x2+2ax+2 =x^2-2x+2=(x-1)^2+1
The opening is upward, and the axis of symmetry is x = 1
Vertex in interval [- 5,5]
When x = 1, there is a minimum value f (1) = 1
|-5-1|＞|5-1|
When x = - 5, there is a maximum value f (- 5) = - 5-1) ^ 2 + 1 = 37

How to find the symmetric axis equation of function y = asin (Wx + F) + B

A just does not affect the expansion of the neglect B is to move up and down does not affect the neglect
Then, the symmetry axis of the whole is calculated, that is, the whole in brackets. Here we should know the SiNx axis of symmetry
Then Wx + F = π / 2 + K π K is an integer, and x =? Is solved, that is, the symmetric axis equation

The answers to questions 1 and 2 of compulsory exercise 1.2 in senior one mathematics, By definition, formula one

For reference:

For rational numbers a, B, define a ⊙ B = 3A + 2B, then calculate [(x + y) ⊙ (X-Y)] ⊙ 3x

[(x+y)⊙（x-y)]⊙3x
=[3（x+y）+2(x-y)]⊙3x
=(3x+3y+2x-2y)⊙3x
=(5x+y)⊙3x
=3(5x+3y)+2×（3x）
=15x+9y+6x
=21x + 9y super simple

Find out the area s of the plane figure surrounded by the curve y = x2 and the straight line x = 1, y = 0, and find the volume of the rotating body obtained by rotating s around the X axis I'm studying

S＝∫(0～1) ydx＝∫(0～1)x^2 dx＝1/3
V＝∫(0～1) πy^2 dx＝∫(0～1) πx^4 dx＝π/5

Through P (1,0), the tangent line of parabola y = radical (X-2) is made, which forms a plane figure with the above parabola and X axis The volume of the body of revolution formed by rotating the figure around the x-axis and y-axis is calculated

The tangent point coordinates P (x0, Y0), (y0-0) / (x0-1) / (x0-1) / (x0-1) = 1 / [2 √ (x0-2)], Y0 = √ (x0-2), (x0-2), [(x0-2)] / (x0-1) = 1 / [2 √ (x0-2)], 2x0-4 = x0-1, x0 = 3, Y0 = 1, tangent point coordinate P (3,1), tangent equation: (y-0) / (x-1) = 1 / 2, y = x / 2-1 / 2, graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics graphics area by curve

It is known that the image of the quadratic function y = ax squared + BX + C and the image of the function y = - 2x + 4x + 1 are symmetric about the x-axis I know the way of thinking, I don't know how to write the format Function wrong, should be: y = - 2x square + 4x + 1

Because the image of the quadratic function y = the square of AX + BX + C and the image of the function y = - 2x ^ 2 + 4x + 1 are symmetric about the X axis,
So - y = - 2x ^ 2 + 4x + 1, y = 2x ^ 2-4x-1

Given the quadratic function f (x) = - x2 + 2 (m-1) x + 2m-m2 (1) if the image passes through the origin, find the value of M, and if the image is symmetric about the y-axis, write the relationship of the function

1. Substituting the origin coordinates (0,0), i.e. x = 0, f (x) = 0, into the function analytic formula F (x) = - x ＾ 2 + 2 (m-1) x + 2m-m ＾ 2
2m-m＾2＝0
m(2-m)=0
ν M = 0 or M = 2
2. Because the symmetry axis of the function is x = - 2 (m-1) / [2 × (- 1)],
When the axis of symmetry is the Y axis (x = 0),
-2(m-1)/〔2×（-1）〕＝0
m-1＝0
m＝1,
In this case, the analytic formula of the function is f (x) = - x ＾ 2 + 1

F (x) = square of root 3sinxcosx + cosx + M (1) Find the minimum positive period and monotone increasing interval (2) if x belongs to [- π / 6, π / 3], the minimum value of F (x) is 2, find the maximum value of F (x) at this time, and point out what value to take F (x) to the maximum value

F (x) = (3) SiNx cosx + cos (s) x + X + m + M = 3 / 2sin2x + 1 / 2 (1 + cos2x) + M = √ 3 / 2 * sin2x + 1 / 2 * cos2x + m + 1 / 2 / 2 = sin (2x + π / 6) + m + 1 / 1 / 2 (1) f (x) minimum positive period T = 2 π / 2 = π from 2K π - π / 2 ≤ 2x + π / 6 ≤ 2K π + π / 2, K ∈ Z, K π - π / 3 ≤ 3 ≤ 3 ≤ 3 ≤ 3 ≤ 3 ≤ 3 ≤ 3 ≤ 3 ≤ 3 ≤ 3 ≤ 2 K π + π + π + π + π + π / 2, K ∈ Z, k x ≤ K π + π / 6, K ∈ Z ν f (x)

Function f (x) = LG (2sinx + 1) range

This is a composite function. From the definition domain of the outer function is the value domain of the inner function and the true number is greater than 0, we know that 2sinx + 1 > 0, and the value range of the inner function - 1 < = 2sinx + 1 < = 3. Therefore, the definition domain of the outer function is 0 < 2sinx + 1 < = 3, and the outer function increases in the definition domain (0,3), so the value range of the outer function is (0, Lg3]