The function f is continuous on the interval [3, 13] with selected values of x and f(x) given in the table below. Use the data in the table to approximate f ′(3.5).
x 3 4 7 10 13
f(x) 2 8 10 12 22
Please HELPP! Thank you
To approximate f'(3.5), we can use the concept of the difference quotient. The difference quotient is defined as:
f'(a) ≈ (f(a + h) - f(a)) / h
For this problem, let's choose h = 0.5. So we want to find f'(3.5), which can be approximated using the values in the table:
f'(3.5) ≈ (f(3.5 + 0.5) - f(3.5)) / 0.5
To compute this approximation, we need to find the values of f(4) and f(3.5):
f(4) = 8
f(3.5) is not given in the table, but we can estimate it by finding the average of the values of f(3) and f(4):
f(3) = 2
f(3.5) ≈ (f(3) + f(4)) / 2 = (2 + 8) / 2 = 5
Now we can substitute these values into the difference quotient:
f'(3.5) ≈ (f(3.5 + 0.5) - f(3.5)) / 0.5
= (f(4) - f(3.5)) / 0.5
= (8 - 5) / 0.5
= 3 / 0.5
= 6
Therefore, the approximate value of f'(3.5) is 6.
To approximate f'(3.5), we can use the concept of the average rate of change. The average rate of change of a function f(x) between two points (x1, f(x1)) and (x2, f(x2)) is given by:
Average rate of change = (f(x2) - f(x1)) / (x2 - x1)
In this case, we can use the points (3, 2) and (4, 8) to calculate the average rate of change between them. Thus, x1 = 3, f(x1) = 2, x2 = 4, and f(x2) = 8.
Average rate of change = (8 - 2) / (4 - 3) = 6 / 1 = 6
Since we are approximating f'(3.5), we can consider the average rate of change as the approximation for f'(3.5), which is approximately 6.
Therefore, the approximate value for f'(3.5) is 6.
why not just use linear interpolation?
3.5 is halfway between 3 and 4, so
f(3.5) is halfway between f(3) and f(4)