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Given the function f (x) = 2cos? X + SiNx, if the definition domain of function f (x) is r, find the value range of function f (x)
When SiNx = 1 / 4, f (x) has the maximum value [f (x)] max = 17 / 8. When SiNx = - 1, f (x) has the minimum value [f (x)] min = - 2-1 + 2 = - 1
If f (x) = a < 2cos ^ 2 (x / 2) + SiNx > + B, a is less than 0, and X belongs to 0 to 180 degrees. The value range of function f is 3 to 4. Find the value of a + B F (x) = a (2cos ^ 2x / 2 + SiNx) + B, a is less than 0, and X belongs to 0 to 180 degrees. The value range of function f is 3 to 4. Find the value of a + B
F(X)=A(CosX+1+SinX)+B
=A(CosX+SinX)+A+B
=2^1/2*A*Sin(x+pi/4)+A+B
When x belongs to [0, PI], sin (x + pi / 4) belongs to [- 2 ^ (- 1 / 2), 1]
When sin (x + pi / 4) = - 2 ^ (- 1 / 2), f (x) = 4
When sin (x + pi / 4) = 1, f (x) = 3
2^1/2*A*[-2^(-1/2)]+A+B=4
2^1/2*A*1+A+B=3
The solution of the equation ^ - 2 is 1 / 2
A+B=5-2^(1/2)
Find the maximum and minimum values of the following functions, and find the set of X whose maximum and minimum values are obtained: 1. Y = SiNx - √ 3cosx 2. Y = SiNx + cosx
The range of the trigonometric function sin α is [- 1,1]
1.y=sinx-√3cosx =2[(1/2)sinx - (√3/2)cosx] =2sin(x - π/3)
The maximum value of the function is 2 and the minimum value of the function is - 2
∵ when the trigonometric function sin α = 1, α = 2K π + π / 2, (K ∈ z)
When the function y takes the maximum value of 2, X - π / 3 = 2K π + π / 2, (K ∈ z)
When x = 2K π + 5 π / 6, (K ∈ z), ymax = 2
∵ when the trigonometric function sin α = - 1, α = 2K π - π / 2, (K ∈ z)
ν - 2 ∈π, where ∈ 2 ∈ π is the smallest
When x = 2K π - π / 6, (K ∈ z), Ymin = - 2
Similarly: 2. Y = SiNx + cosx = √ 2 [(√ 2 / 2) SiNx + (√ 2 / 2) cosx] = √ 2Sin (x + π / 4)
The maximum value of the function is √ 2, and the minimum value of the function is - √ 2
When x = 2K π + π / 4, (K ∈ z), ymax = √ 2
When x = 2K π - 3 π / 4, (K ∈ z), Ymin = - √ 2
What is the tangent equation of the curve y = cosx at the point (π / 3,1 / 2); what is the normal equation?
y'=-sinx
x=π/3,y'=-√3/2
This is the tangent slope
From point oblique
y-1/2=-√3/2(x-π/3）
3√3x+6y-3-π√3=0
Normal vertical tangent, slope 2 √ 3 / 3
So it's 12 √ 3x-18y-4 π√ 3 + 9 = 0
Find tangent equation and normal equation of curve y ^ 3 + y ^ 2 = 2x at point (1,1)
Deriving x
3y^2*y'+2y*y'=2
y'=2/(3y^2+2y)=2/(3+2)=2/5
So the tangent slope is 2 / 5
So Y-1 = (2 / 5) (x-1)
2x-5y+3=0
The normal is perpendicular to the tangent, so the slope = - 5 / 2
So, 2x-7
Find tangent equation and normal equation of curve y = x ^ 3-1 at point (1,0)
∵y=x^3-1
∴y'=3x²
Then the slope of (1,1 ') of (y) = 1
The tangent equation is: y-0 = 3 (x-1), that is, 3x-y-3 = 0
The slope of normal is - 1 / 3
The normal equation is: y-0 = - 1 / 3 * (x-1), that is, x + 3y-1 = 0
Find the normal equation and tangent equation of the curve What are the tangent equation and normal equation of curve {x = 2t-t ^ 2, y = 3t-t ^ 3 at t = 0? I calculated that x = 0, y = 0, DX / dt = x '= (2t-t ^ 2) = 2-2t dy/dt=y'=(3t-t^3)=3-3^3 dy/dx=(3-3t^3)/(2-2t) How to calculate the normal equation and tangent equation finally
When t = 0, x = 0, y = 0, so the tangent point is (0,0)
When DX / dt = 2-2t, t = 0, DX / dt = 2
When dy / dt = 3-3t ^ 2, t = 0, dy / dt = 3
Therefore, when t = 0, dy / DX = 3 / 2
So, the tangent equation is y = 3x / 2, and the normal equation is y = - 2x / 3
The derivative of y = 1 / root x is obtained by definition. What is the slope of tangent at y = 1 / root x? Please explain in detail
In this paper, we discuss the problem of (x + X + X + x) - 1 / / / [(x + X + x) - x] / [(x + X + x) - x]] = LIM (△ x) [((x + X + △ x) / (x + △ x) - 1 / / [x] / (x + △ x) - 1 / / [x] / (x + (x + △ x)] / [(x + (x + (x) + x)] / [(x + X + x), X]] = LIM (△ x → 0) [1 / √ x (x + X + (x)] [(x ^ 2 ^ 2)] [(x ^ 2 ^ 2, 2)] [(x ^ 2, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, + X △ x) - (x + △ x)] / △ x where LIM (△ x
What is the tangent slope of the curve y = 1 / (root x) at x = 9? The answer is - 1 / 54
The slope of the tangent is the value of derivative at the point ~ y = 1 / root sign x = x ^ (- 1 / 2), ^ is the power, x ^ (- 1 / 2) is the - 1 / 2 power of X, y '= (- 1 / 2) x ^ (- 3 / 2) = (- 1 / 2) x ^ (- 3 / 2). Therefore, when x = 9, the slope of the tangent at this point k = y' = (- 1 / 2) × 9 ^ (- 3 / 2) = (- 1 / 2) × (1 / 2) × (1 / 2) = (- 1 / 2) × (1 / 2) = (- 1 / 2) × (1
The curve f (x) = the slope of the tangent line at point (1,2) of the radical x is
f'(x)=1/(2√x)
f'(1)=1/2
RELATED INFORMATIONS
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