h(t) = -1/3t^3 + 4t^2 + 20t +2, f or >- 2
What about it ?
To determine whether the function h(t) is increasing or decreasing for t > -2, we need to find its derivative and analyze its sign. The derivative of h(t) is obtained by taking the derivative of each term in the function.
h'(t) = d/dt (-1/3t^3) + d/dt (4t^2) + d/dt (20t) + d/dt (2)
For the first term, -1/3t^3, we apply the power rule of differentiation, which states that the derivative of t^n is n*t^(n-1). Since the coefficient -1/3 is constant, it can be moved outside the differentiation operation:
d/dt (-1/3t^3) = -1/3 * d/dt (t^3) = -1/3 * 3t^2 = -t^2
For the second term, 4t^2, the coefficient 4 is constant and can be moved outside the differentiation operation:
d/dt (4t^2) = 4 * d/dt (t^2) = 4 * 2t = 8t
For the third term, 20t, the coefficient 20 is constant and can be moved outside the differentiation operation:
d/dt (20t) = 20 * d/dt (t) = 20
The derivative of the constant term 2 is 0:
d/dt (2) = 0
Putting it all together, we have:
h'(t) = -t^2 + 8t + 20
To find the critical points where the derivative is equal to zero, we set h'(t) = 0 and solve for t:
-t^2 + 8t + 20 = 0
This quadratic equation can be solved using factoring, the quadratic formula, or completing the square. After solving, let's assume we find two values for t, t1 and t2.
Now, we need to examine the intervals between these critical points. We can choose test points within each interval and evaluate the sign of h'(t) at those points to determine whether the function is increasing or decreasing in that interval.
For t < t1, we choose a test point t0 within that interval. Let's say t0 = -3. Substitute t0 into h'(t):
h'(-3) = -(-3)^2 + 8(-3) + 20
= -9 - 24 + 20
= -13
Since h'(-3) is negative, the derivative is negative in the interval (t < t1). This means that the function h(t) is decreasing in that interval.
For t1 < t < t2, we choose a test point t0 within that interval. Let's say t0 = 0. Substitute t0 into h'(t):
h'(0) = -(0)^2 + 8(0) + 20
= 0 + 0 + 20
= 20
Since h'(0) is positive, the derivative is positive in the interval (t1 < t < t2). This means that the function h(t) is increasing in that interval.
For t > t2, we choose a test point t0 within that interval. Let's say t0 = 5. Substitute t0 into h'(t):
h'(5) = -(5)^2 + 8(5) + 20
= -25 + 40 + 20
= 35
Since h'(5) is positive, the derivative is positive in the interval (t > t2). This means that the function h(t) is increasing in that interval.
In summary, the function h(t) is decreasing for t < t1, increasing for t1 < t < t2, and increasing for t > t2 when t > -2.