To find the average rate of change of the function h(x) = x^2 + 3x + 2 on the interval -3 < x < a, we need to calculate the slope between the two endpoints of the interval.
The slope between two points (x1, y1) and (x2, y2) is given by (y2 - y1) / (x2 - x1).
In this case, the two endpoints of the interval are (-3, h(-3)) and (a, h(a)).
Using the function h(x) = x^2 + 3x + 2, we can calculate the values of h(-3) and h(a):
h(-3) = (-3)^2 + 3(-3) + 2 = 4
h(a) = a^2 + 3a + 2
The slope between these two points is:
(-1) = (h(a) - h(-3)) / (a -(-3))
(-1) = (a^2 + 3a + 2 - 4) / (a + 3)
Simplifying the equation:
(-1) = (a^2 + 3a - 2) / (a + 3)
Multiplying both sides of the equation by (a + 3):
(-1)(a + 3) = (a^2 + 3a - 2)
Simplifying further:
-a - 3 = a^2 + 3a - 2
Rearranging the equation:
a^2 + 4a - 1 = 0
To solve this quadratic equation, we can use the quadratic formula:
a = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 4, and c = -1.
a = (-(4) ± sqrt((4)^2 - 4(1)(-1))) / (2(1))
a = (-4 ± sqrt(16 + 4)) / 2
a = (-4 ± sqrt(20)) / 2
a = (-4 ± 2sqrt(5)) / 2
a = -2 ± sqrt(5)
So, the possible values of a that satisfy the condition of the average rate of change being -1 units/unit on the interval -3 < x < a are:
a = -2 + sqrt(5) or a = -2 - sqrt(5)
However, since the interval is given as -3 < x < a, the value of a must be greater than -3. Therefore, the only valid solution is:
a = -2 + sqrt(5)
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<\ is -1 units/unit
1 answer