1. Determine the slope of the tangent to the graph of y=4/x at (1,4)

my answer is -4

2.What is the slope of the line perpendicular to y=(x-2)/(x+2) at (2,0)

my answer is -4

3. For which value of x is the derivative of f(x)=(x-1)^10(x+2)^5 equal to 0?

my answer is 1

1st two are corrrect

#3.
my derivative was
f ' (x) = 15(x+1)(x-1)^9 (x+2)^4
= 0 for

x = -1, 1, -2

verify at Wolfram
http://www.wolframalpha.com/input/?i=plot+y+%3D+%28x-1%29%5E10%28x%2B2%29%5E5

1. Well, it seems like you've got the right answer! The slope of the tangent to the graph of y = 4/x at (1, 4) is indeed -4. Great job!

2. Ah, perpendicular lines, always in a constant state of conflict! You're correct again, the slope of the line perpendicular to y = (x-2)/(x+2) at (2, 0) is -4. Those lines just can't help but be opposites!

3. Hmm, finding where the derivative goes through a "zero party"? Sounds like a blast! And you're right, my friend, the derivative of f(x) = (x-1)^10(x+2)^5 equals 0 at x = 1. Enjoy the festivities!

1. To find the slope of the tangent to the graph of y = 4/x at (1,4), we need to find the derivative of the function at that point. The derivative of y = 4/x can be found using the quotient rule, which states that if f(x) = u(x)/v(x), then the derivative of f(x) is (u'(x)v(x) - u(x)v'(x))/[v(x)]^2.

In this case, u(x) = 4 and v(x) = x. Therefore, u'(x) = 0 and v'(x) = 1.

Applying the quotient rule, the derivative of y = 4/x is (0*x - 4*1)/(x^2), which simplifies to -4/x^2.

To find the slope of the tangent at (1,4), substitute x = 1 into the derivative expression:
slope = -4/(1)^2 = -4/1 = -4

Therefore, the slope of the tangent to the graph of y = 4/x at (1,4) is -4.

2. To find the slope of the line perpendicular to y = (x-2)/(x+2) at (2,0), we first need to find the derivative of the function at that point. The derivative of y = (x-2)/(x+2) can be found using the quotient rule, as explained in the previous explanation.

Applying the quotient rule, the derivative of y = (x-2)/(x+2) is [(1*(x+2) - (x-2)*1)/[(x+2)^2].

To find the slope of the line perpendicular to y = (x-2)/(x+2), we can take the negative reciprocal of the derivative at (2,0). Let's call the derivative D(x).

Slope of the line perpendicular = -1/D(2)

Substitute x = 2 into the derivative expression:
D(2) = [(1*(2+2) - (2-2)*1)/[(2+2)^2] = [(4 - 0)/16] = 4/16 = 1/4

Therefore, the slope of the line perpendicular to y = (x-2)/(x+2) at (2,0) is -1/(1/4) = -4.

3. To find the value of x for which the derivative of f(x) = (x-1)^10(x+2)^5 is equal to 0, we need to find the critical points of the function. A critical point occurs where the derivative is equal to 0.

Let's call the derivative of f(x) D(x).

To find the critical points, we need to find the values of x where D(x) = 0.

D(x) = 0 when [(10(x-1)^9(x+2)^5) + (5(x-1)^10(x+2)^4)(1)] = 0

Simplifying the equation and factoring out common terms, we get:

(10(x-1)^9(x+2)^4) [(x-1) + (x+2)] = 0

For this equation to equal 0, either (10(x-1)^9(x+2)^4) = 0 or [(x-1) + (x+2)] = 0.

1) (10(x-1)^9(x+2)^4) = 0

Setting each factor to 0, we have:

x-1 = 0 --> x = 1
x+2 = 0 --> x = -2

2) [(x-1) + (x+2)] = 0

Simplifying, we get:

2x + 1 = 0 --> x = -1/2

Therefore, the values of x for which the derivative of f(x) = (x-1)^10(x+2)^5 is equal to 0 are x = 1, x = -2, and x = -1/2.

To determine the slope of the tangent to a graph or the slope of a line perpendicular to a given equation, you can use calculus and algebraic methods.

1. To find the slope of the tangent to the graph of y = 4/x at (1,4):
- Take the derivative of the function y = 4/x with respect to x. The derivative of 4/x is -4/x^2.
- Substitute x = 1 into the derivative to find the slope at that point: -4/1^2 = -4.
Therefore, the slope of the tangent to the graph of y = 4/x at (1,4) is -4.

2. To find the slope of the line perpendicular to y = (x-2)/(x+2) at (2,0):
- Determine the derivative of the function y = (x-2)/(x+2). Take the derivative using the quotient rule: [(x+2)(1) - (x-2)(1)] / (x+2)^2.
- Simplify the expression: (2x + 4) / (x+2)^2.
- Substitute x = 2 into the derivative to find the slope at that point: (2(2) + 4) / (2+2)^2 = 8/16 = 1/2.
The slope of the tangent at (2,0) is 1/2.
The slope of a line perpendicular to this tangent will have a negative reciprocal slope, which is -2. Therefore, the slope of the line perpendicular to y = (x-2)/(x+2) at (2,0) is -2.

3. To find the value of x when the derivative of f(x) = (x-1)^10(x+2)^5 equals 0:
- Take the derivative of the function f(x) = (x-1)^10(x+2)^5 with respect to x.
- Apply the product rule and simplify the expression.
- Set the derivative equal to zero and solve the equation for x.
By solving the equation, you will find the value(s) of x for which the derivative of f(x) equals 0. In this case, the value is x = 1.