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AP Calculus: AB
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f(-2) = -5
f(-2) = -5
All reals
all x not equal to 0, 1
It is 2 because g(x) is always 2 (constant).
2^3 -3(2^2) -2*2 + 5 = -3
x^3 + 1
1/2 and -1
x + 2 = sqrt(4x), x + 2 = 2 * sqrt(x), (x+2)^2 = 4x, so x^2 + 4x + 4-4x = 0, so x^2 + 4x = 0 so x^2 = -4x which means they don't intersect.
x since for an inverse function, inserting it as an argument into the original function and vice versa always returns x.
3 * pi / 2
0
-1
1
nonexistent
nonexistent
5x^4 * tan(x) + x^5 * sec^2(x)
(-(3x+1) - 3*(2-x)) / (3x+1)^2
(((e^x - 1) / (e^x)) * (e^x - 1)/x - (e^x / x)) / (e^x - 1)^2
(sec x tan x + sec^2 (x)) / (sec x + tan x)
cos(1/x) (-x^(-2))
e^(-x) -2sin(2x) -e^-x * cos(2x)
y' = 8*x^(-1/2), so y'' = -4x^(-3/2)
2x + 2yy' = 0, so x + yy' = 0, y' = -y/x
Infinity
cos(xy)*(xy' + y) = 1, solve for y'.
3y^2y' - (x2yy' + y^2) = 0, so at y = 2, y' = 1/2
-xe^-x
Vasilios S.
Carnegie Mellon University
AP Calculus: AB Tutor
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