Question

Let f(x)= ⎩ ⎨ ⎧ ​ 0 5 −2 0 ​ if x<−3 if −3≤x<0 if 0≤x<5 if x≥5 ​ and g(x)=∫ −3 x ​ f(t)dt Determine the value of each of the following: (a) g(−7)= (b) g(−2)= (c) g(1)= (d) g(6)= (e) The absolute maximum of g(x) occurs when x= and is the value It may be helpful to make a graph of f(x) when answering these questions.

Answer 1

The function g(x) is determined by integrating the piecewise function f(x) from -3 to x. The values of g(x) are determined by the periods where f(x) contributes to the area under the curve. The absolute maximum of g(x) occurs at x = 0.

Step-by-step explanation:

The student is asking for the evaluation of the function g(x), which is defined as the integral of another piecewise function f(x) from -3 to x. To solve this, we'll calculate the integral of each section of the piecewise function, taking into account the intervals given for f(x). Given the nature of f(x) and its intervals, we can establish the value of g(x) by integrating the relevant section from -3 to x.

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The absolute maximum of g(x) would occur at the point where g(x) transitions from increasing to decreasing. Since g(x) is the integral of f(x), this point is where f(x) transitions from positive to zero or negative, which corresponds to x = 0 since from -3 to 0, f(x) is 0 and from 0 onwards it is not positive.

Answer 2

Step-by-step explanation:

To determine the values of g(x) for the given inputs, we need to evaluate the integral of f(t) with respect to t over the specified intervals. Let's calculate each value step by step:

(a) g(-7):

Since x < -3, the integral evaluates to zero:

g(-7) = ∫[-3, -7] f(t) dt = 0

(b) g(-2):

Since -3 ≤ x < 0, the integral evaluates from -3 to -2:

g(-2) = ∫[-3, -2] f(t) dt

To calculate this integral, we need to split it into two parts, since f(t) has different values for -3 ≤ t < 0 and 0 ≤ t < -2:

g(-2) = ∫[-3, 0] f(t) dt + ∫[0, -2] f(t) dt

For the first part, -3 ≤ t < 0, f(t) = -2:

∫[-3, 0] f(t) dt = ∫[-3, 0] -2 dt = -2[t] from -3 to 0 = -2(0 - (-3)) = -2(3) = -6

For the second part, 0 ≤ t < -2, f(t) = 0:

∫[0, -2] f(t) dt = ∫[0, -2] 0 dt = 0[t] from 0 to -2 = 0 - 0 = 0

Combining the two parts:

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

Therefore, g(-2) = -6.

(c) g(1):

Since 0 ≤ x < 5, the integral evaluates from -3 to 1:

g(1) = ∫[-3, 1] f(t) dt

Again, we need to split the integral into two parts:

g(1) = ∫[-3, 0] f(t) dt + ∫[0, 1] f(t) dt

For the first part, -3 ≤ t < 0, f(t) = -2:

∫[-3, 0] f(t) dt = ∫[-3, 0] -2 dt = -2[t] from -3 to 0 = -2(0 - (-3)) = -2(3) = -6

For the second part, 0 ≤ t < 1, f(t) = 0:

∫[0, 1] f(t) dt = ∫[0, 1] 0 dt = 0[t] from 0 to 1 = 0 - 0 = 0

Combining the two parts:

g(1) = ∫[-3, 0] f(t) dt + ∫[0, 1] f(t) dt = -6 + 0 = -6

Therefore, g(1) = -6.

(d) g(6):

Since x ≥ 5, the integral evaluates from -3 to 5:

g(6) = ∫[-3, 5] f(t) dt

For the entire interval -3 ≤ t < 5, f(t) = 0:

∫[-3, 5] f(t) dt = ∫[-3, 5] 0 dt = 0[t] from -3 to 5 = 0 - 0 = 0

Therefore, g(6) = 0.

(e) The absolute maximum of g(x):

To find the absolute maximum of g(x), we need to examine the values of g(x) over the entire domain of x, which is from negative infinity to positive infinity. By analyzing the graph of f(x), we can see that the maximum value of g(x) occurs at x = 5, where the function jumps from 0 to -2.

Thus, the absolute maximum of g(x) occurs when x = 5, and its value is -2.