I neeed your help terrriblyyyyy (calculus) -- Please HELP!?

Question

#1

If f(x)= sqrt(x^3 + 5x +121) * (x^2 +x +11) then, f'(0)=?

Note that the square root does not extend over (x^2 +x +11).
It's the square root of (x^3 + 5x +121) multiplied by (x^2 +x +11).


#2
if y is a differentiable function of x, then the slope of the curve xy^2 -2y + 4y^3 = 6 at the point where y = 1 is what?


#3
Find all open intervals on which the following function is decreasing:
f(x) = x / (x^2 + x -2)


It's crucial that I see your work (how you got the answer).
Thanks a million. You people are awsome.

Answer 1

Well, I see you've used the correct order of operations this time. That's good - it shows that you're improving.

#1: Apply the product and chain rules to find f'(x), then evaluate at 0:

f'(x) = d(√(x³+5x+121))/dx (x²+x+11) + (2x+1)√(x³+5x+121)

f'(x) = 1/(2√(x³+5x+121)) * d(x³+5x+121)/dx * (x²+x+11) + (2x+1)√(x³+5x+121)

f'(x) = 1/(2√(x³+5x+121)) * (3x²+5) * (x²+x+11) + (2x+1)√(x³+5x+121)

Now evaluating :

f'(0) = 1/(2√121)*5*11 + √121 = 5/2 + 11 = 27/2

#2: First, evaluate the function at y=1 to get the value of x:

xy^2 -2y + 4y^3 = 6
x - 1 + 4 = 6
x = 3

Now, implicitly differentiate both sides:

d(xy^2 -2y + 4y^3)/dx = d(6)/dx
y² + 2xy dy/dx - 2 dy/dx + 12y² dy/dx = 0

Solve for dy/dx:

y² + (2xy - 2 + 12y²) dy/dx = 0
(2xy - 2 + 12y²) dy/dx = -y²
dy/dx = -y²/(2xy - 2 + 12y²)

Now evaluate at (3, 1):

dy/dx = -1/(6 - 2 + 12) = -1/16

#3: First, find the derivative:

f'(x) = (x²+x-2 - (2x+1)x)/(x²+x-2)² (using the quotient rule)

Simplifying:

f'(x) = (x²+x-2 - 2x² - x)/(x²+x-2)²
f'(x) = (-x²-2)/(x²+x-2)²

Now we must find where f'(x) < 0

(-x²-2)/(x²+x-2)² < 0
-x²-2 < 0 (we know (x²+x-2)² is nonnegative, and if the expression on the left is defined, it is also nonzero. Thus it is positive and we can multiply by it without changing the inequality)
-x² < 2

Now, since x is a real number, this expression will always be true. Thus, if f'(x) is even defined, x is decreasing. Now f'(x) is defined provied that (x²+x-2)²≠0. So:

(x²+x-2)²≠0
(x-1)²(x+2)²≠0
x≠1 and x≠-2

Which makes sense, since those are vertical asymptotes of the original function. Thus, the function is decreasing on:

(-∞, -2) ∪ (-2, 1) ∪ (1, ∞)

Answer 2

#1
If f(x)= sqrt(x^3 + 5x +121) * (x^2 +x +11)
f'(x) = (2x + 1)√(x^3 + 5x +121) + (1/2)(x^2 +x +11)(3x + 5)/√(x^3 + 5x +121)
f"(0) = (0 + 1)√(0 + 0121) + (1/2)(0 + 0 +11)(0 + 5)/√(0 + 0 +121)
f"(0) = √121 + (1/2)(11)(5)/√121
f"(0) = 11 + (1/2)(5)
f"(0) = 13.5

#2
xy^2 -2y + 4y^3 = 6
x - 2 + 4 = 6
x = 4

2*4dy + dx - 2dy +12dy = 0
(8 - 2 +12)dy = - dx
dy/dx = -1/18

#3
f(x) = x / (x^2 + x - 2) = x/((x + 2)(x - 1))
f(x) = 1 / (x + 1 - 2/x)
The function increases from x = -∞ to x = -2, decreases from -2 < x< 1, a second asymptote, and increases again from 1 < x < ∞

Answer 3

1. plug in 0 for all the x's
f(0) = sqrt(0+0+121)*(0+0+11) = sqrt(121)*11 = 11*11 = 121