Estimate f′(−3) where f(x) = x^3 + 8x^2 − 10x − 1 to within one decimal place by using a small enough interval.
f′(−3) ≈
It is an estimation. You can use the first order approximations such as
f'(x) = (f(x+δ)-f(x))/δ
or a symmetrical formula, such as
(f(x+δ)-f(x-δ)/(2*δ)
δ can be a small number such as 0.1, 0.01,0.001,etc.
To estimate f′(−3), we can use the concept of the derivative. The derivative of a function gives us the rate at which the function is changing at a particular point. In this case, we are asked to estimate the derivative at x = −3 for the function f(x) = x^3 + 8x^2 − 10x − 1.
To estimate the derivative, we can use the concept of the average rate of change. The average rate of change of a function over a small interval is equal to the slope of the secant line between the two points on the function.
To find the estimate, we need to choose a small interval around x = −3. Let's choose a small interval of, for example, h = 0.1, which means we will consider the points x = −3 and x = −2.9.
We can calculate the average rate of change by finding the difference in the function values divided by the difference in the x-values:
Average rate of change = (f(−2.9) - f(−3)) / (−2.9 - (−3))
Now let's calculate the function values:
f(−2.9) = (−2.9)^3 + 8(−2.9)^2 − 10(−2.9) − 1
Finally, substitute the function values into the average rate of change formula and calculate the estimate.
f′(−3) ≈ average rate of change (rounded to one decimal place)
By repeating this process with smaller intervals, we can improve the accuracy of our estimate.