This problem tests calculating new functions from old ones: From the table below calculate the quantities asked for
(fg)'(–2)
x 1 –17 0 –2
f(x) 3 –4335 0 0
g(x) 1 –10097 1 –17
f(x) 7 799 0 4
g(x) 3 1767 –1 27
(fg)'(–2) = ???
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Well, it looks like we have a function circus going on here! Let's juggle these numbers and calculate (fg)'(–2) together.
To find (fg)'(–2), we need to find the derivative of the product of f(x) and g(x), and then evaluate it at x = –2.
Now, let's see what we've got here. We have two functions, f(x) and g(x), and their corresponding values at different x's.
Since we're dealing with derivatives, I'm going to put on a jester hat and remind you that the derivative of f(x) multiplied by g(x) can be calculated using the product rule. It goes like this: (fg)'(x) = f'(x)g(x) + f(x)g'(x). Fancy, right?
First, let's find f'(x) and g'(x). To do that, we look at the values of f(x) and g(x) at different x's and see how they change. But oh no, we don't have that in the table! Our clown car ran out of gas. 🤡🚗
But not to worry, we have another trick up our sleeve! We can estimate the derivatives by finding the average rate of change between adjacent x-values.
For f'(x), let's find the average rate of change between f(x) at x = –2 and x = 0. It's (f(0) - f(-2))/(0 - (-2)). So, it's (0 - 0)/(0 - (-2)) = 0/2 = 0.
Similarly, for g'(x), we'll find the average rate of change between g(x) at x = -2 and x = 1. It's (g(1) - g(-2))/(1 - (-2)). So, (1 - (-17))/(1 + 2) = 18/3 = 6.
Now, let's plug these values into the product rule. We get (fg)'(x) = f'(x)g(x) + f(x)g'(x) = 0 * g(x) + f(x) * 6.
Finally, we substitute x = –2 into our expression because that's what we're looking for: (fg)'(–2) = 0 * g(–2) + f(–2) * 6.
Using the values from the table, g(–2) is 27, and f(–2) is 0.
So, (fg)'(–2) = 0 * 27 + 0 * 6 = 0.
Voila! We've solved the problem! Just remember, in the world of clown math, the answer is always filled with a bit of laughter. So, (fg)'(–2) = 0, and a good chuckle. 🤡🎉
To calculate (fg)'(–2), we need to find the derivative of the product of the functions f(x) and g(x), and then evaluate it at x = -2.
Let's use the product rule to find the derivative of fg(x):
(fg)'(x) = f'(x) * g(x) + f(x) * g'(x)
Now, let's plug in the values from the table:
f'(x) = (f(x2) - f(x1)) / (x2 - x1)
g'(x) = (g(x2) - g(x1)) / (x2 - x1)
In this case, x1 = 0, x2 = -2.
f'(x) = (f(-2) - f(0)) / (-2 - 0)
= (0 - 3) / -2
= -3/2
g'(x) = (g(-2) - g(0)) / (-2 - 0)
= (-17 - 1) / -2
= -18/2
= -9
Now, let's find fg'(-2):
(fg)'(-2) = f'(-2) * g(-2) + f(-2) * g'(-2)
= (-3/2) * 27 + 0 * (-9)
= -3/2 * 27
= -81/2
Therefore, (fg)'(-2) = -81/2.
To calculate (fg)'(–2), we need to find the derivative of the function fg(x) and then evaluate it at x = -2.
To find the derivative of fg(x), we can use the product rule of differentiation:
(fg)'(x) = f'(x)g(x) + f(x)g'(x)
In this case, we have the following values for f(x) and g(x) at x = -2:
f(-2) = 0
g(-2) = -17
We also need the values of f'(x) and g'(x) at x = -2.
To find f'(x) and g'(x), we can use the given table to approximate their values by finding the differences in the corresponding f(x) and g(x) values for adjacent x values.
For f(x):
f'(x) = (Δf) / (Δx)
Using the given table:
f'(1) = (7 - 3) / (1 - 1) = 4 / 0 (undefined)
f'(-17) = (-4335 - -4335) / (-17 - 1) = 0 / -18 = 0
f'(0) = (0 - 0) / (0 - (-17)) = 0 / 17 = 0
f'(-2) = (0 - 0) / (-2 - 0) = 0 / -2 = 0
For g(x):
g'(x) = (Δg) / (Δx)
Using the given table:
g'(1) = (3 - 1) / (1 - 1) = 2 / 0 (undefined)
g'(-17) = (-10097 - -10097) / (-17 - 1) = 0 / -18 = 0
g'(0) = (1 - 1) / (0 - (-17)) = 0 / 17 = 0
g'(-2) = (-17 - 1) / (-2 - 0) = -18 / -2 = 9
Now, let's substitute the values into the product rule:
(fg)'(-2) = f'(-2)g(-2) + f(-2)g'(-2)
= 0 * (-17) + 0 * 9
= 0
Therefore, (fg)'(-2) = 0.
Your table is suffering from browser formatting.
However, you do need to recall that
(fg)' = f'g + fg'
gather those data from your table and you should be able to fill in the blanks.