Asked by Ke$ha
f is a continuous function with a domain [−3, 9] such that
f of x equals 3 for x between negative 3 and 0 including negative 3, equals negative 1 times x plus 3 for x between 0 and 6 inclusive, and equals negative 3 for x greater than 6 and less than or equal to 9
and let g of x equals the integral from negative 2 to x of f of t, dt.
On what interval is g increasing? Justify your answer.
For 0 ≤ x ≤ 6, express g(x) in terms of x. Do not include +C in your final answer.
f of x equals 3 for x between negative 3 and 0 including negative 3, equals negative 1 times x plus 3 for x between 0 and 6 inclusive, and equals negative 3 for x greater than 6 and less than or equal to 9
and let g of x equals the integral from negative 2 to x of f of t, dt.
On what interval is g increasing? Justify your answer.
For 0 ≤ x ≤ 6, express g(x) in terms of x. Do not include +C in your final answer.
Answers
Answered by
Steve
why all those messy words? You appear to be saying
f(x) =
3 for -3 <= x < 0
-x+3 for 0 <= x <= 6
-3 for 6 < x <= 9
g is increasing where f is positive: [-2,0)U(3,6]
g(x) = ∫[-2,0] 3 dt + ∫[0,x] -t+3 dt
= 6 + (-x^2/2 + 3x)+3 for 0<=t<=6
f(x) =
3 for -3 <= x < 0
-x+3 for 0 <= x <= 6
-3 for 6 < x <= 9
g is increasing where f is positive: [-2,0)U(3,6]
g(x) = ∫[-2,0] 3 dt + ∫[0,x] -t+3 dt
= 6 + (-x^2/2 + 3x)+3 for 0<=t<=6
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