Compulsory exercise 1.2b group 1~

Compulsory exercise 1.2b group 1~

1.()1{p｜-5《=p

Compulsory exercise 1. 2B group 2, the answer to the third question, the best picture

If x belongs to (- 2.5, - 2), y = - 3, if x belongs to (- 2.5, - 2), y = - 3[-

The third problem of group 2B, Today Can you type it for me? I've been there. If I can't, I'll give you 10 points Sorry, I don't understand

f（ x)={-3,-2.5

(1) y = 3x (2) y = 8 (3) y = minus 4x + 5 (4) y = x squared - 6x + 7 Draw the function picture below and tell the coordinates of the four questions (1)y=3x (2) Y = 8 / X (3) Y = minus 4x + 5 (4) Y = square of X - 6x + 7

A straight line passing through the origin (0,0) (1,3)
Is a hyperbola, a three quadrant (1,8) (8,1) (- 1, - 8) (- 8, - 1)
A straight line (1,1) (2, - 3)
It's a parabola. The vertex passes through the point (1,2) (- 1,12) at (3, - 2)

The function y = f (x) = asin (Wx + b), (a > 0,4 > W > 0,0)

When x = 0, y = √ 3, that is √ 3 = 2sinb SINB = √ 3 / 2 b = π / 3 or 2 π / 3 when y = 0, x = 2 π / 9 = 0 sin (2 π w / 9 + b) = 0 sin (2 π w / 9 + b) = 0 2 π w / 9 + B = π or 2 π w / 9 + B

The function y = asin (Wx + φ) The function y = asin (Wx + φ) (where a > 0, w > 0, | φ)|

T=2π/w=π ===>w=2
The ordinate of the highest point is 3 / 2 = = = > A = 3 / 2
An equation of symmetry axis of image is x = π / 6
Then 2 * (π / 6) + φ = k π + π / 2 = = > φ = k π + π / 6
When k = 0, φ = π / 6
So the analytic formula of (2) sin = (2 x π)
Increasing interval 2K π - π / 2 < = 2x + π / 6 < = 2K π + π / 2
K π - π / 3 < = x < = k π + π / 6 is obtained
Function subtraction interval 2K π + π / 2 < = 2x + π / 6 < = 2K π + 3 π / 2
K π + π / 6 < = x < = k π + 2 π / 3 is obtained

Find the area of the plane figure enclosed by curve y = x ^ 2 and curve y ^ 2 = X

Intersection point of two curves (0,0), (1,1)
The integral interval is [0,1] and it is known that y 2 = x is above y = x 2
→∫(√x - x² )dx
The next step is the calculation

Find the area of the plane figure enclosed by the curve y = x ^ 2-3 and the straight line y = 2x

The intersection point of y = x ^ 2 and the straight line y = 2x are: (0,0) and (2,4) taking DX as the differential element, the integral formula is as follows: Area s = integral (0,4) (2x-x ^ 2) DX (because y = 2x is above y = x ^ 2 on (0,2)) s = integral (2,0) (2x-x ^ 2) DX = (x ^ 2-x ^ 3 / 3) | (2,0) the integral is determined by Newton Leibniz theorem and S = 4-8 / 3 = 4 / 3

Find the area of the plane figure enclosed by the curves y = x ^ 3 (the cube of x) and y = 2x

Y = x ^ 3 and y = 2x intersect at three points (- √ 2,0), (0,0), (√ 2,0)
Area = 2 * integral (0 to √ 2) [2x-x ^ 3] DX = 2 * (0 to √ 2) [x ^ 2-x ^ 4 / 4] = 2 [(√ 2) ^ 2 - (√ 2) ^ 4 / 4] = 2

Find the area of plane figure enclosed by curve y = x square + x-3 and y = 2x-1

Simultaneous y = x ^ 2 + x-3 and y = 2x-1
X = 2 or x = - 1
Let f '(x) = x ^ 2 + x-3 - (2x-1) = x ^ 2-x-2
f(x)=(1/3)x^3-(1/2)x^2-2x
Calculus of area s = f '(x) = / F (2) - f (- 1) / = (9 / 2)