Testing Your Limits
lim x→0 (3x-1)/x
ln3
Find f'(x)
f(x)=(4x+5)3
12(4X+5)2
find dy/dx given x2+y2=9
-x/y
∫(x2+1)2dx=
1/5x5+2/3x3+x+C
If lim x→3 f(x)=7, which of the following must be true?
I. f is continuous at x=3
II. f is differentiable at x=3
III f(3)=7
None because f(3) is not stated
Is this function differentiable at x=3?
f(x)=3x2-30, x ≥ 3
f(x)=x3-10x, x < 3
No
d/dx sec-1(3x)
3/(|3x|√(9x2 - 1))
If x2+xy+y3=0, then, in terms of x and y, dy/dx is
-(2x+y)/(x+3y2)
∫ 0 to π of (cosx+1)dx
π
lim n→∞ (3n3-5n)/(n3-2n2+1)
3
lim x→∞ (3x5-6x)/ex
0
Approximate f(2.2) using the line tangent to f(x)=√(2+x) at x=2
2.05 OR 41/20
If tan(xy)=x, then dy/dx is
(cos2(xy)-y)/x OR (1-ysec2(xy))/(xsec2(xy))
∫sec2xdx
tanx+C
The acceleration of a particle moving along the x-axis at time t is given by a(t)=6t-2. If the velocity is 25 when t=3 and the position is 10 when t=1, then the position x(t)=
x(t)=t3-t2+4t+6
What type of discontinuity?
g(x)=x2-5, x < 3
g(x)=9-2x, x ≥ 3
Jump Discontinuity
If f(x)=√(x2-4) and g(x)=3x-2, then the derivative of f(g(x)) at x=3 is
7/√5 or (7√5)/5
If x+2xy-y2=2, then at the point (1, 1), dy/dx is
Does not exist at (1,1)
∫ 2 to 3 of x(x2+1)dx
75/4
For what value of x does the function f'(x)=(x-2)(x-3)2 have a relative maximum?
7/3
lim x→0 sin(3x)/(2-√(4-x))
12
Let f(x)=(2x+1)3, and let g(x) be the inverse function of f. Given that f(0)=1, what is g'(1)?
1/6
If cos(x2)=xey, then, dy/dx is
(-2xsin(x2)-ey)/(xey)
If ∫ 1 to 10 of f(x)dx=4 and ∫ 10 to 3 of f(x)dx=7, then ∫ 1 to 3 of f(x) is
11
A particle moves along a line so that at time t, where 0≤t≤π, its position is given by s(t)=-4cos(t)-(t2/2)+10. What is the velocity of the particle when its acceleration is zero?
v(1.31812)=2.55487