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.
Why did the function go to therapy? Because it had some issues with its domain! Let's find out on what interval g is increasing and then express g(x) for 0 ≤ x ≤ 6.
To determine where g is increasing, we need to find the intervals where f(x) is positive. So, let's analyze the three parts of f(x).
1. For x between -3 and 0 (inclusive), f(x) = 3. This means that f(x) is positive in this interval.
2. For x between 0 and 6 (inclusive), f(x) = -1(x + 3). Since x + 3 is always positive in this interval, f(x) will be negative. So, g(x) will be decreasing on this interval.
3. For x between 6 and 9 (inclusive), f(x) = -3. This means that f(x) is negative in this interval.
Based on this analysis, we can conclude that g is increasing on the interval -3 ≤ x ≤ 0.
Now let's express g(x) for 0 ≤ x ≤ 6. We know that g(x) is the integral of f(t) from -2 to x.
For 0 ≤ x ≤ 6:
∫[0 to x] f(t) dt = ∫[-2 to 0] f(t) dt + ∫[0 to x] f(t) dt
= 0 + ∫[0 to x] f(t) dt
= ∫[0 to x] 3 dt
Since f(t) is a constant 3 in this interval, we can simply integrate the constant term. Thus, g(x) for 0 ≤ x ≤ 6 is:
g(x) = 3x (No +C term needed because we are only evaluating the integral from 0 to x)
Keep laughing and math-ing!
To determine on what interval g(x) is increasing, we need to analyze the derivative of g(x). If the derivative is positive, it means that g(x) is increasing on that interval.
Let's calculate the derivative of g(x) using the Second Fundamental Theorem of Calculus:
g'(x) = f(x)
From the given information:
For x ∈ [-3, 0], f(x) = 3
For x ∈ [0, 6], f(x) = -1 * x + 3
For x ∈ (6, 9], f(x) = -3
Now, let's break down the intervals and analyze the sign of g'(x) in each interval:
1. For x ∈ [-3, 0]:
Here, f(x) = 3, which is positive. So, g'(x) = f(x) = 3. Hence, g(x) is increasing on [-3, 0].
2. For x ∈ [0, 6]:
Here, f(x) = -1 * x + 3, and the slope of the linear function is negative (-1). Hence, g'(x) = f(x) = -1 * x + 3. As the slope is negative, g(x) would be decreasing on [0, 6].
3. For x ∈ (6, 9]:
Here, f(x) = -3, which is a constant. Hence, g'(x) = f(x) = -3. As the slope is negative, g(x) would be decreasing on (6, 9].
Therefore, g(x) is increasing only on the interval [-3, 0].
To determine on what interval the function g(x) is increasing, we need to find the values of x for which the derivative of g(x) is positive.
First, let's find the derivative of g(x) using the fundamental theorem of calculus. According to the fundamental theorem of calculus, if F(x) is the antiderivative of f(x), then the derivative of the integral from a to x of f(t) dt is simply f(x).
In this case, f(x) is given as follows:
For x ≤ 0: f(x) = 3
For 0 ≤ x ≤ 6: f(x) = -1(x + 3)
For x > 6: f(x) = -3
So, we can write g(x) as:
For x ≤ 0: g(x) = ∫[−2, x] 3 dt = 3(x - (-2)) = 3(x + 2)
For 0 ≤ x ≤ 6: g(x) = ∫[−2, 0] 3 dt + ∫[0, x] (-1)(t + 3) dt = 3(x + 2) + ∫[0, x] -(t + 3) dt
For x > 6: g(x) = ∫[−2, 0] 3 dt + ∫[0, 6] -(t + 3) dt + ∫[6, x] -3 dt = 3(x + 2) + [-(x + 3) * (x - 0)] + [-3(x - 6)].
Now let's find the derivative of g(x) for the different intervals.
For x ≤ 0: g'(x) = d/dx [3(x + 2)] = 3
For 0 ≤ x ≤ 6: g'(x) = d/dx [3(x + 2) - (x + 3) * (x - 0)] = 3 - (x + 3)
For x > 6: g'(x) = d/dx [3(x + 2) + (-(x + 3) * (x - 0)) - 3(x - 6)] = 3 - (x + 3) - 3
Now, let's analyze the intervals:
For x ≤ 0: g'(x) = 3 > 0, so g(x) is increasing on this interval.
For 0 ≤ x ≤ 6: g'(x) = 3 - (x + 3), the derivative is negative for all values in this interval, so g(x) is not increasing on this interval.
For x > 6: g'(x) = 3 - (x + 3) - 3 = -x - 3, the derivative is negative for all values in this interval, so g(x) is not increasing on this interval.
Therefore, the function g(x) is increasing only for x ≤ 0.
Next, we need to express g(x) in terms of x for 0 ≤ x ≤ 6:
For 0 ≤ x ≤ 6: g(x) = 3(x + 2) + ∫[0, x] -(t + 3) dt
To calculate the integral, we can simplify -(t + 3) as -t - 3:
g(x) = 3(x + 2) + ∫[0, x] -(t + 3) dt
= 3(x + 2) - ∫[0, x] (t + 3) dt
= 3(x + 2) - [(1/2)t^2 + 3t] evaluated from 0 to x
= 3(x + 2) - [(1/2)x^2 + 3x] - [(1/2)(0)^2 + 3(0)]
= 3(x + 2) - (1/2)x^2 - 3x
Therefore, for 0 ≤ x ≤ 6, g(x) = 3(x + 2) - (1/2)x^2 - 3x.
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