∫ (x2+3) = ∫x2 + ∫3(1/3)X3 + 3X + C
lim (x3 + x2 + 3x + 3) / (x4 + x3 + 2x + 2)x > -1From the cave of the ancient stone tablets, we cleared away several feet of cobwebs and unearthed"l'Hospital's" rule: If substitution of the limit results in ( 0/0 ), then the limit is equal to the(limit of the derivative of the numerator) divided by (limit of the derivative of the denominator).(3x2 + 2x + 3) / (4x3 + 3x2 + 2) evaluated at (x = -1) is:(3 - 2 + 3) / (-4 + 3 + 2) = 4 / 1 = 1
2
2.5
x2 + 6x + 1 = 0 x2 + 6x + 9 = 8 (x + 3)2 = 8 x + 3 = ± 2√2 x + 3 = -3 ± 2√2 x ∈ {-3 - 2√2, -3 + 2√2}
The anti-derivative of X2 plus X is the same as the anti-derivative of X2 plus the anti-derivative of X. The anti derivative of X2 is X3/3 plus an integration constant C1 The anti derivative of X is X2/2 plus an integration constant C2 So the anti-derivative of X2+X is (X3/3)+(X2/2)+C1+C2 The constants can be combined and the fraction can combined by using a common denominator leaving (2X3/6)+(3X2/6)+C X2/6 can be factored out leaving (X2/6)(2X+3)+C Hope that helps
Following the correct order of operations: derivative of x^2 + 6/2 = derivative of x^2 +3, which equals 2x
∫ (x2+3) = ∫x2 + ∫3(1/3)X3 + 3X + C
f'(x) = (x2 - x)1/21. Add one to the exponent2. Divide step 1 by the exponent plus one= (x2 - x)1/2+1____________1/2 + 1= (x2 - x)3/2_________ 3/2= (2/3) (x2 - x)3/2 + CIt may also be written:[2(x2 - x)3/2 / 3] + CCheck:f(x) = (2/3) (x2 - x)3/2Using the chain rule:f'(x) = (2/3) (3/2) (x2 - x)3/2-1= (x2 - x)1/2
-1
lim (x3 + x2 + 3x + 3) / (x4 + x3 + 2x + 2)x > -1From the cave of the ancient stone tablets, we cleared away several feet of cobwebs and unearthed"l'Hospital's" rule: If substitution of the limit results in ( 0/0 ), then the limit is equal to the(limit of the derivative of the numerator) divided by (limit of the derivative of the denominator).(3x2 + 2x + 3) / (4x3 + 3x2 + 2) evaluated at (x = -1) is:(3 - 2 + 3) / (-4 + 3 + 2) = 4 / 1 = 1
2
2.5
To find the extreme value of the parabola y = x2 - 4x + 3 ...(1) Take the derivative of the equation.y = x2 - 4x + 3y' = 2x - 4(2) Set the derivative = 0 and solve for x.y' = 2x - 40 = 2x - 42x = 4x = 4/2x = 2(3) Plug this x value back into the original equation to find the associated y coordinate.x = 2y = x2 - 4x + 3y = (2)2 - 4(2) + 3y = 4 - 8 + 3y = -1So the vertex is at (2, -1).
x2 + 6x + 1 = 0 x2 + 6x + 9 = 8 (x + 3)2 = 8 x + 3 = ± 2√2 x + 3 = -3 ± 2√2 x ∈ {-3 - 2√2, -3 + 2√2}
x3 + 2x2 + 3x + 6 = x2(x + 2) + 3(x + 2) = (x + 2)(x2 + 3)
x2 + 5x + 6 = (x + 2)(x + 3)