英文微积分1.an equcation of the lone tangent to the graph of y=cos(2x) at x= π/4 isA.y-1= -(x-π/4)B,y-1= -2(x-π/4)C.y=2(x-π/4)D.y= -(x-π/4)E.y= -2(x=π/4)2.d/dx cos^2(x^3)=3.0.5∫e^0.5 dt=4.the graph of y=3x^4-16x^3+24x^2+48 is concave dow

英文微积分1.an equcation of the lone tangent to the graph of y=cos(2x) at x= π/4 isA.y-1= -(x-π/4)B,y-1= -2(x-π/4)C.y=2(x-π/4)D.y= -(x-π/4)E.y= -2(x=π/4)2.d/dx cos^2(x^3)=3.0.5∫e^0.5 dt=4.the graph of y=3x^4-16x^3+24x^2+48 is concave dow
英文微积分
1.an equcation of the lone tangent to the graph of y=cos(2x) at x= π/4 is
A.y-1= -(x-π/4)
B,y-1= -2(x-π/4)
C.y=2(x-π/4)
D.y= -(x-π/4)
E.y= -2(x=π/4)
2.d/dx cos^2(x^3)=
3.0.5∫e^0.5 dt=
4.the graph of y=3x^4-16x^3+24x^2+48 is concave down for
A.x0
C.x< -2 or x> -2/3
D.x2
这是我们老师发的某年的AP EXAM试题

英文微积分1.an equcation of the lone tangent to the graph of y=cos(2x) at x= π/4 isA.y-1= -(x-π/4)B,y-1= -2(x-π/4)C.y=2(x-π/4)D.y= -(x-π/4)E.y= -2(x=π/4)2.d/dx cos^2(x^3)=3.0.5∫e^0.5 dt=4.the graph of y=3x^4-16x^3+24x^2+48 is concave dow
1.An equation of the line tangent to the graph of y=cos(2x) at x= π/4 is
A.y-1= -(x-π/4)
B,y-1= -2(x-π/4)
C.y=2(x-π/4)
D.y= -(x-π/4)
E.y= -2(x=π/4)
【Solution】：
dy/dx = -[sin(2x)](2) = -2sin(2x)
dy/dx|(x=π/4) = -2sin(π/2) = -2
Let tangent line be y = -2x + c
Sub x=π/4 and y = cos(π/2) = 0 sub into y = -2x + c,we get c = π/2
So,tangent line is y = -2x + π/2
Ans :None of the above
(typo happened with your ans)
2.d/dx cos^2(x^3)=d[cos²x³]/dx = -2(cosx³)(sinx³)(3x²) = -3x²sin²(2x³)
3.0.5∫e^0.5 dt= 0.5(e^0.5)t + c (maybe mistyping)
0.5∫e^(0.5t) dt= (e^0.5t) + c
4.the graph of y=3x^4-16x^3+24x^2+48 is concave down for
A.x0
C.x< -2 or x> -2/3
D.x2
【Solution】：
y = 3x⁴- 16x³ + 24x² + 48
dy/dx = 12x³ - 48x² + 48x = 12x(x² - 4x + 4) = 12x(x - 2)²
Let dy/dx = 0
We have x₁= 0；x₂= 2
d²y/dx² = 36x² - 96x + 48 = 12(3x² - 8x + 4) = 12(x-2)(3x-2)
When 2/3 < x

Suppose f(x) is a continuous function with positive values and
f(0)=1. If for any x>0, the length of the curve y=f(x) over the interval [0,x] is always equal to the area of the region below th...

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Suppose f(x) is a continuous function with positive values and
f(0)=1. If for any x>0, the length of the curve y=f(x) over the interval [0,x] is always equal to the area of the region below this curve and above the x-axis, find the equation of this curve.
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