Asked by Jennifer
Determine the value of 'a' so that the average rate of change of the function h(x)=x^2+3x+2 on the interval -3<=x<=a is -1
Answers
Answered by
Damon
h(a) = a^2 + 3 a + 2
h(-3)= 9 - 9 + 2
h(a) -h(-3) = a^2 + 3 a
so (a^2 + 3 a) /(a+3) = -1
a^2 + 3 a = -a - 3
a^2 + 4 a + 4 = 0
(a+2)(a+2) = 0
a = -2
h(-3)= 9 - 9 + 2
h(a) -h(-3) = a^2 + 3 a
so (a^2 + 3 a) /(a+3) = -1
a^2 + 3 a = -a - 3
a^2 + 4 a + 4 = 0
(a+2)(a+2) = 0
a = -2
Answered by
Jennifer
Awesome Thanks!!
Answered by
Damon
You are welcome :)
Answered by
Damon
a^2 + 4 a = -a -3
a^2 + 4 a + 3 = 0
a = [ -4 +/- sqrt (16-12)]/2
a = -2 +/- 1
a = -1 or -3
a^2 + 4 a + 3 = 0
a = [ -4 +/- sqrt (16-12)]/2
a = -2 +/- 1
a = -1 or -3
Answered by
Jennifer
Got it! thanks for the correction.
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