Friend of mine needs an explanation of this problem.
1 answer:
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Answer:
I) f + g = 10/3
II) 4f + 2g = 20/3
III) f = 2 and g = 4/3
Step-by-step explanation:
From the chart,
P = 25
q = 40
The total number of one centimeter lines in the first n diagrams is given by the expression
2/3n^3 + fn^2 + gn.
When n = 1, the total number of line = 4. So,
2/3(1)^3 + f(1)^2 + g(1) = 4
2/3 + f + g = 4
Make f+g the subject of formula
f + g = 4 - 2/3
f + g = (12 - 2)/3
f + g = 10/3 ......(1)
When n = 2
Total number of line = 12
2/3(2)^3 + f(2)^2 + g(2) = 12
2/3×8 + 4f + 2g = 12
16/3 + 4f + 2g = 12
4f + 2g = 12 - 16/3
4f + 2g = (36 - 16)/3
4f + 2g = 20/3 ......(2)
(iii) To find the values of f and g, solve equation 1 and 2 simultaneously
f + g = 10/3 × 2
4f + 2g = 32/3
2f + 2g = 20/3
4f + 2g = 32/3
- 2f = - 12/3
f = 12/6
f = 2
Substitutes f in equation 1
f + g = 10/3
2 + g = 10/3
g = 10/3 - 2
g = (10 - 6)/3
g = 4/3
Answer:
(-5, 0) ∪ (5, ∞)
Step-by-step explanation:
I find a graph convenient for this purpose. (See below)
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When you want to find where a function is increasing or decreasing, you want to look at the sign of the derivative. Here, the derivative is ...
f'(x) = 4x^3 -100x = 4x(x^2 -25) = 4x(x +5)(x -5)
This has zeros at x=-5, x=0, and x=5. The sign of the derivative will be positive when 0 or 2 factors have negative signs. The signs change at the zeros. So, the intervals of f' having a positive sign are (-5, 0) and (5, ∞).
