Step 1: Calculate the function values at x=−1 and x=3.
Given f(x)=−2x3.
For x=−1:
f(−1)=−2(−1)3=−2(−1)=2
For x=3:
f(3)=−2(3)3=−2(27)=−54
Step 2: Apply the formula for the average gradient.
The average gradient is given by x2−x1f(x2)−f(x1).
Average gradient=3−(−1)f(3)−f(−1)=3+1−54−2=4−56
Average gradient=−14
The average gradient between x=−1 and x=3 is −14.
Step 1: Write down the definition of the derivative from first principles.
f′(x)=limh→0hf(x+h)−f(x)
Step 2: Find f(x+h).
Given f(x)=−2x3.
f(x+h)=−2(x+h)3
Expand (x+h)3:
(x+h)3=x3+3x2h+3xh2+h3
Substitute this back into f(x+h):
f(x+h)=−2(x3+3x2h+3xh2+h3)=−2x3−6x2h−6xh2−2h3
Step 3: Calculate f(x+h)−f(x).
f(x+h)−f(x)=(−2x3−6x2h−6xh2−2h3)−(−2x3)
f(x+h)−f(x)=−2x3−6x2h−6xh2−2h3+2x3
f(x+h)−f(x)=−6x2h−6xh2−2h3
Step 4: Divide by h and simplify.
hf(x+h)−f(x)=h−6x2h−6xh2−2h3
Factor out h from the numerator:
hh(−6x2−6xh−2h2)=−6x2−6xh−2h2
Step 5: Take the limit as h→0.
f′(x)=limh→0(−6x2−6xh−2h2)
Substitute h=0:
f′(x)=−6x2−6x(0)−2(0)2
f′(x)=−6x2
The derivative f′(x) from first principles is −6x2.
Step 1: Use the result from 1.1.2, which is f′(x)=−6x2.
Substitute x=−1 into the derivative function:
f′(−1)=−6(−1)2
f′(−1)=−6(1)
f′(−1)=−6
The value of f′(−1) is −6.
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