Let f(x)=1.81.7x. Estimate the derivative f′(1.1) to within two decimal places by using a small enough interval.
f′(1.1)≈
3.0043
1.81.7x makes no sense
To estimate the derivative f′(1.1) of the function f(x) = 1.81.7x, we can use the concept of the derivative as the slope of the tangent line to the function at a specific point.
To find the derivative, we can calculate the difference quotient using a small enough interval around the point x = 1.1.
Let's choose a small interval, such as h = 0.01. We will calculate the average rate of change of f(x) over this interval, which will give us an approximation of the derivative.
f′(1.1) ≈ (f(1.1 + h) - f(1.1)) / h
Plugging in the values:
f(1.1 + 0.01) = 1.81.7(1.1 + 0.01)
= 1.81.7(1.11)
= 1.94.7
f(1.1) = 1.81.7(1.1)
= 1.81.7(1.1)
= 1.89.7
Substituting back into the difference quotient:
f′(1.1) ≈ (1.94.7 - 1.89.7) / 0.01
Simplifying the numerator:
f′(1.1) ≈ 0.05.700 / 0.01
Dividing:
f′(1.1) ≈ 5.700
Therefore, f′(1.1) ≈ 5.70 to two decimal places.
To estimate the derivative f′(1.1) using a small enough interval, we can use the definition of the derivative. The derivative of a function represents its instantaneous rate of change at a specific point.
The definition of the derivative is given by:
f′(x) = lim(h → 0) [f(x + h) - f(x)] / h
To estimate the derivative at x = 1.1, we need to evaluate the function f(x) at x = 1.1 and also slightly above and below this point.
Let's choose a small value for h, such as h = 0.01. We will use this value to calculate the estimate of the derivative.
Calculating f(1.1 + 0.01):
f(1.11) = 1.7 * 1.11 = 1.887
Calculating f(1.1 - 0.01):
f(1.09) = 1.7 * 1.09 = 1.853
Now, we can use these values to estimate the derivative f′(1.1):
f′(1.1) ≈ [f(1.1 + 0.01) - f(1.1 - 0.01)] / (0.01)
≈ (1.887 - 1.853) / (0.01)
≈ 0.034 / 0.01
≈ 3.4
Therefore, the estimate of the derivative f′(1.1) is approximately 3.4 to two decimal places.