Let f(x)=1x^3+2.5x+3. In this problem you will estimate f'(-3) by computing the average rate of change of f(x)over a small interval. Round your answers to three decimal places.
1. First choose the interval of length .1 with -3 being the left endpoint of the interval. Then f'(-3)¡Ö
2. Now let the interval have length .01. Then f'(-3)¡Ö
3. Finally let the interval have .001. Then f'(-3)¡Ö
To approximate the derivative of f(x) at -3, we can compute the average rate of change of f(x) over small intervals centered at -3.
1. For an interval of length 0.1 with -3 as the left endpoint, we compute f(-3 + 0.1) - f(-3) and divide by the length of the interval. Let's calculate it step by step:
- Evaluate f(-3 + 0.1):
f(-3 + 0.1) = 1(-3 + 0.1)^3 + 2.5(-3 + 0.1) + 3
= 0.729 + (-7.25) + 3
= -3.521
- Evaluate f(-3):
f(-3) = 1(-3)^3 + 2.5(-3) + 3
= -27 + (-7.5) + 3
= -31.5
- Calculate the average rate of change:
f'(-3) ≈ (f(-3 + 0.1) - f(-3)) / (0.1)
≈ (-3.521 - (-31.5)) / 0.1
≈ 28.979 / 0.1
≈ 289.79
So, f'(-3) ≈ 289.79.
2. For an interval of length 0.01, the process is the same as above, just with smaller intervals:
- Evaluate f(-3 + 0.01):
f(-3 + 0.01) = 1(-3 + 0.01)^3 + 2.5(-3 + 0.01) + 3
≈ -3.000029
- Evaluate f(-3):
f(-3) ≈ -31.5 (as calculated before)
- Calculate the average rate of change:
f'(-3) ≈ (f(-3 + 0.01) - f(-3)) / (0.01)
≈ (-3.000029 - (-31.5)) / 0.01
≈ 28.499 / 0.01
≈ 2849.9
So, f'(-3) ≈ 2849.9.
3. Similarly, for an interval of length 0.001:
- Evaluate f(-3 + 0.001):
f(-3 + 0.001) = 1(-3 + 0.001)^3 + 2.5(-3 + 0.001) + 3
≈ -3.000000000027
- Evaluate f(-3):
f(-3) ≈ -31.5 (as calculated before)
- Calculate the average rate of change:
f'(-3) ≈ (f(-3 + 0.001) - f(-3)) / (0.001)
≈ (-3.000000000027 - (-31.5)) / 0.001
≈ 28.5 / 0.001
≈ 28500
So, f'(-3) ≈ 28500.
By using smaller intervals, we can approach the true value of f'(-3), which is the derivative of f(x) at -3.