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f(x)=|9-x|; closed interval=[7,10]

2007-05-02 07:25:22 · 5 answers · asked by kaitie 1 in Science & Mathematics Mathematics

5 answers

Keep doing your homework using Yahoo Answers and you won't ever learn anything.

2007-05-02 07:30:33 · answer #1 · answered by Lyvy 4 · 0 0

First draw a graph of the function, then integrate it over the interval [7,10]. Finally divide the result by the length of the interval [7,10], which is 10-7=3.

2007-05-02 07:30:35 · answer #2 · answered by Anonymous · 0 0

Since we are dealing with an absolute value term, you have to ensure that where it would normally be negative, it is positive. For this term, The value at x=7 is 2, and drops linearly to 0 at x=9. Then it increases linearly to 1 at x=10. In effect you have two triangles. Find the area of both and divide by 3.

2007-05-02 07:32:20 · answer #3 · answered by cattbarf 7 · 0 0

permit F(x) be the critical of the functionality f(x). the known fee of the functionality is defined as: favg = (F(b) - F(a))/(b - a) the place b and a define the era that we are attracted to looking the advise fee over. For f(x) = |8 - x| over the era 5 <= x <= 9, we've fairly a concern, so we could desire to split the critical into 2 aspects. permit: g(x) be the functionality for the era 5 <= x <= 8. considering we would like g(x) to equivalent f(x) over this variety, g(x) = 8 - x G(x) be the critical of g(x) for the era 5 <= x <= 8 h(x) be the functionality for the era 8 <= x <= 9. considering we would like g(x) to equivalent f(x) over this variety, g(x) = x - 9 H(x) be the critical for the era 8 <= x <= 9 G(x) = ?(8 - x)dx G(x) = (8x - 0.5x^2) H(x) = ?(x - 8)dx H(x) = (0.5*x^2 - 8x) comparing G over its era: G = G(8) - G(5) G = (8*8 - 0.5*8^2) - (8*5 - 0.5*5^2) G = (sixty 4 - 0.5*sixty 4) - (40 - 0.5*25) G = (sixty 4 - 32) - (40 - 12.5) G = 32 - 27.5 G = 4.5 comparing H over its era: H = H(9) - H(8) H = (0.5*9^2 - 8*9) - (0.5*8^2 - 8*8) H = (0.5*80 one - seventy two) - (0.5*sixty 4 - sixty 4) H = (40.5 - seventy two) - (32 - sixty 4) H = -31.5 - (-32) H = 0.5 going returned to the favg functionality: favg = (F(9) - F(5))/(9 - 5) favg = (F(9) - F(5))/4 Now F(9) - F(5) is the area decrease than the curve for the era 5 <= x <= 9. that's comparable to G + H. favg = (G + H)/4 favg = (4.5 + 0.5)/4 favg = 5/4 favg = a million.25

2016-10-14 09:02:54 · answer #4 · answered by Anonymous · 0 0

and x belongs to Q? Z? R?

2007-05-02 07:30:14 · answer #5 · answered by kiki 3 · 0 0

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