What is the instantaneous slope of y =8/x at x = -3?
What is the average rate of change of y with respect to x over the interval [-3, 4] for the function y = 2x + 2?
The surface area, S, of a sphere of radius r feet is S = S(r) = 4πr2. Find the instantaneous rate of change of the surface area with respect to the radius r at r = 2.
I am a little confused here can anyone show the work and steps please.
i have no idea
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To find the instantaneous slope of a function, you need to find the derivative of the function and then evaluate it at the given value of x. Here's how to do it:
1. Take the derivative of the function y = 8/x with respect to x. The derivative of 8/x can be found using the power rule for differentiation:
d/dx (8/x) = -8/x^2
2. Now, substitute the given value of x = -3 into the derivative:
-8/(-3)^2 = -8/9
Therefore, the instantaneous slope of y = 8/x at x = -3 is -8/9.
To find the average rate of change of a function over an interval, you need to find the difference in the function values (y-values) divided by the difference in the input values (x-values) over that interval. Here's how to do it:
1. Substitute the starting x-value and the ending x-value into the function y = 2x + 2 to find the corresponding y-values:
For x = -3: y = 2(-3) + 2 = -4
For x = 4: y = 2(4) + 2 = 10
2. Now, calculate the difference in the y-values and the difference in the x-values:
Δy = 10 - (-4) = 14
Δx = 4 - (-3) = 7
3. Finally, divide the difference in the y-values by the difference in the x-values to get the average rate of change:
Average rate of change = Δy/Δx = 14/7 = 2
Therefore, the average rate of change of y with respect to x over the interval [-3, 4] for the function y = 2x + 2 is 2.
To find the instantaneous rate of change of a function with respect to a variable at a specific point, you need to find the derivative of the function with respect to that variable and then evaluate it at the given point. Here's how to do it:
1. Take the derivative of the function S = 4πr^2 with respect to r. The derivative can be found using the power rule for differentiation:
dS/dr = d/dx (4πr^2) = 8πr
2. Now, substitute the given value of r = 2 into the derivative:
dS/dr at r = 2 = 8π(2) = 16π
Therefore, the instantaneous rate of change of the surface area with respect to the radius r at r = 2 is 16π.
y = 8/x
y' = -8/x^2
now plug in x = -3
the avg rate of change is
change in y
------------------
change in x
So, that would be
(y(4)-y(-3))/(4-(-3))
Now just plug in the numbers