f is a continuous function with a domain [−3, 9] such that

f(x)= 3 , -3 ≤ x < 0
-x+3 , 0 ≤ x ≤ 6
-3 , 6 < x ≤ 9

and let g(x)= ∫ f(t) dt where a=-2 b=x

On what interval is g increasing? Justify your answer.

For 0 ≤ x ≤ 6, express g(x) in terms of x.

Just so people know, he is right on the second question and wrong on the first one.

Well, it seems like we have a function party going on here! Let's break it down step by step, shall we?

First, we have the function f(x) given to us with different pieces for different intervals. I see f(x) wearing a 3-piece suit for -3 ≤ x < 0, then rocking the straight-line look of -x+3 for 0 ≤ x ≤ 6, and finally donning a casual -3 outfit for 6 < x ≤ 9. Quite the fashionista!

Now, for g(x), we have an integral to deal with. It's like the DJ of the party, mixing things up. We need to use the function f(t) and integrate it from -2 to x. So, we're basically finding the area under the curve of f(t) from -2 to x.

To determine when g(x) is increasing, we need to look at its derivative. If the derivative is positive, g(x) is increasing; if it's negative, g(x) is decreasing. If it's neutral, well, I guess g(x) is just chilling and not going anywhere.

To express g(x) for 0 ≤ x ≤ 6, we need to find the integral of f(t) from -2 to x specifically for this interval. Remember that f(x) changes depending on the interval.

So, let's put our calculators to work and find the integral of f(t) for 0 ≤ x ≤ 6 using the given functions for f(x). And remember, when it comes to math, no clowning around!

...Calculating...

Alrighty then! After some integration magic, we find that g(x) for 0 ≤ x ≤ 6 is:

g(x) = ∫[0 to x] f(t) dt = ∫[0 to x] (-t + 3) dt

Now you can go ahead and find that integral to express g(x) in terms of x. Just take it one step at a time, and don't forget to add that "+ C" at the end.

As for determining when g(x) is increasing, you'll need to analyze the derivative of g(x) and look for intervals where it's positive. That's where the party is really heating up!

So, get your math mojo on and have fun finding the answer. Remember, even math problems deserve a little laughter along the way!

To determine on what interval g(x) is increasing, we need to find the derivative of g(x) with respect to x and analyze its sign.

Given that g(x) = ∫ f(t) dt and a = -2, b = x, we can find g(x) using the fundamental theorem of calculus.

To find g(x) on the interval 0 ≤ x ≤ 6, we need to evaluate the integral for the respective intervals of f(t).

On the interval -3 ≤ x < 0:
∫(from -2 to x) 3 dt = 3(x - (-2)) = 3(x + 2) = 3x + 6

On the interval 0 ≤ x ≤ 6:
∫(from -2 to 0) 3 dt + ∫(from 0 to x) (-t + 3) dt = 3(0 - (-2)) + (-1/2)(x^2 - 0^2) + 3(x - 0) = 6 + (-1/2)x^2 + 3x

On the interval 6 < x ≤ 9:
∫(from -2 to 0) 3 dt + ∫(from 0 to 6) (-t + 3) dt + ∫(from 6 to x) (-3) dt = 6 + (-1/2)(6^2 - 0^2) + 3(6 - 0) + (-3)(x - 6) = 18 - (1/2)x^2

Now, let's find the derivative of g(x) with respect to x to determine where it is increasing.

For 0 ≤ x ≤ 6, the derivative of g(x) can be found by taking the derivative of the expression 6 + (-1/2)x^2 + 3x:

g'(x) = 0 - (1/2)(2x) + 3 = -x + 3

To find where g(x) is increasing, we need to find the values of x such that g'(x) > 0.

Setting -x + 3 > 0, we solve for x:

-x + 3 > 0
x < 3

Therefore, on the interval 0 ≤ x ≤ 6, g(x) is increasing since the derivative g'(x) = -x + 3 > 0 for x < 3.

To determine the intervals on which g(x) is increasing, we need to understand the behavior of the function f(x) within the given interval [−3, 9].

First, let's find the integral of f(t) with respect to t over the interval [a, b], where a = -2 and b = x:

∫[a,b] f(t) dt

Since f(t) is given as three different values for different intervals, we need to split the integration into three parts, corresponding to the respective intervals:

1. For -3 ≤ t < 0: ∫[-3,0] 3 dt = 3t
2. For 0 ≤ t ≤ 6: ∫[0,t] (-t+3) dt = -0.5t^2 + 3t
3. For 6 < t ≤ 9: ∫[6,9] -3 dt = -3t

Now, let's express g(x) in terms of x, specifically for 0 ≤ x ≤ 6. Since the integration is split into different intervals, we only need to focus on the interval 0 ≤ x ≤ 6:

g(x) = ∫[-2,x] f(t) dt

Since the interval of integration is within the interval defined for f(t) where -x+3 is valid, we only need to consider the integral from the start of this interval (0) to x:

g(x) = ∫[0,x] (-t+3) dt

Integrating this expression will help us find g(x) in terms of x:

g(x) = ∫[0,x] (-t+3) dt
= [-0.5t^2 + 3t] evaluated from 0 to x
= -0.5x^2 + 3x - (-0.5(0)^2 + 3(0))
= -0.5x^2 + 3x

Now, to determine the intervals on which g(x) is increasing, we need to analyze the derivative of g(x). The derivative of g(x) will help us understand the sign of g'(x) and identify the intervals where it is positive.

Taking the derivative of g(x) with respect to x:

g'(x) = d/dx (-0.5x^2 + 3x)
= -x + 3

We can see that g'(x) = -x + 3 is a linear function. To find where g'(x) > 0 (positive), we need to find the values of x that satisfy the inequality:

-x + 3 > 0

Solving this inequality, we get:

x < 3

Therefore, g(x) is increasing for x values less than 3, or in interval (-∞, 3).

In conclusion, g(x) is increasing on the interval (-∞, 3).

g(x) is increasing where f(x) is positive, since f(x) = g'(x)

for x in [-3,0)
g(x) = ∫[-2,x] 3 dt = 3t[-2,x] = 3x+6

for x in [0,6]
g(x) = ∫[-2,0] 3 dx + ∫[0,x] (-t+3) dt
= 3t[-2,0] + (-t^2/3+3t)[0,x]
= 6 + (-x^2/2 + 3x) - 0
= -x^2/2 + 3x + 6