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Tangent equation and normal equation of X + 1 at (1.2) under y = root sign
Y '= 1 / (2 x)
∴k=f'(1)=1/2
The tangent equation is: Y-2 = 1 / 2 (x-1), that is, y = 1 / 2x + 3 / 2
The normal slope is: - 2
The normal equation is: Y-2 = - 2 (x-1), that is: y = - 2x + 4
The tangent slope of the curve y = root x + 1 / 1 at point (0,1) is A. Negative half B. One half C. Double root sign (x + 1) 1 / 3 D direction is the negative of C
Point on curve
So that's the cut-off point
y=(x+1)^(-1/2)
(- 2) = (- 2) = (- 2)
x=0,y'=-1/2
So the slope k = - 1 / 2
Option a
The tangent slope of the curve y = root x at point (1.1) is?
X ^ (- 0.5) / 2. The slope is 1 / 2 when x = 1 is substituted
Curve y = SiNx sinx+cosx−1 2 at point m (π) The slope of tangent at 4,0) is () A. −1 Two B. 1 Two C. − Two Two D. Two Two
∵y＝sinx
sinx+cosx−1
Two
∴y'=cosx(sinx+cosx)−(cosx−sinx)sinx
(sinx+cosx)2
=1
(sinx+cosx)2
y'|x=π
4=1
(sinx+cosx)2|x=π
4=1
Two
Therefore, B
Find the tangent slope of curve y = 1 / root x at point (1,1)
Slope = - 1 / 2
Given that the curve y = 5 times the root sign 2x, find the tangent equation on the curve parallel to the straight line y = 2X-4, and find the tangent equation that passes through point P (0,5) and is tangent to the curve Take a look. It's 2x under y = 5 times root, not y = 5 times root X
1. (1 / 2) y '= (5 / 2) (2x) ^ (1 / 2) y' = (5 / 2) (2x) ^ (- 1 / 2) * (2x) '= 5 / √ (2x) is parallel, then tangent slope = 25 / √ (2x) = 2x = 25 / 8y = 25 / 2, so it's 8x-4y + 25 = 02, set point (a, 5 √ 2a) tangent slope (a, 5 √ 2a) tangent slope 5 / √ (2a) so Y-5 √ (2a) = 5 / √ (2a) * (x-a) over p5-5 √ (2a) = 5 / √ (2a) * (2a) √ (2a) * (2a) (2a) has passed p5- 5-5 √ (2a) = 5 / √ (2a) * (2a) * (2a) (2a) / (2a) - 2
The slope of the tangent line at a point (π / 3, root 2 / 3, root 3) on the graph of function y = SiNx is
K=y'(π/3)=cosπ/3=1/2
The known function f (x) = asin (2x + φ) (a > 0,0
(1) Obviously, a = 1
The solution of √ 3 / 2 = sin (π / 3 + φ) is obtained by introducing the point m (π / 6, √ 3 / 2) into the solution to obtain φ = π / 3
So f (x) = sin (2x + π / 3)
Obviously, its value range is [- 1,1]
(2) According to 2K π + π / 2
It is known that the maximum value of the function f (x) = asin (x + φ) (a > 0,0 < φ < π, X ∈ R) is 2 and its image passes through point m (π / 3,1) (1) If Tan = 3 and the function g (x) = f (x + α) + F (x + α - π / 2) (x ∈ R), the image is symmetric with respect to the straight line x = x, and the value of TaNx is obtained,
tanα=3
It is known that the maximum value of the function y = asin (ω x + Pai / 4) is 2 and the minimum positive period is 8. If the abscissa of two points P and Q in the image of function FX are 2 and 40 respectively, the area of triangle poq is calculated
One
A> 0, w > 0
fx=Asin(wx+pai/4) （A>0,w>0)
The maximum value is 2, a = 2,
The minimum positive period is 8,
From 2 π / w = 8, w = π / 4
∴f(x)=2sin(π/4*x+π/4)
Two
When x = 2, f (2) = 2Sin (π / 2 + π / 4) = √ 2
When x = 4, f (4) = 2Sin (π + π / 4) = - √ 2
∴P(2,√2),Q(4,-√2)
Midpoint m (3,0) of line PQ
The area of the triangle poq
S=SΔPOM+SΔQOM
=3×√2×1/2+3×√2×1/2
=3√2
RELATED INFORMATIONS
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