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Sergeu [11.5K]
2 years ago
8

Find the most general antiderivative of the function. (check your answer by differentiation. use c for the constant of the antid

erivative. ) g(t) = 8 t t2 t
Mathematics
1 answer:
katovenus [111]2 years ago
8 0

The most general antiderivative of the given function g(t) is (8t + t³/3 + t²/2 + c).

The antiderivative of a function is the inverse function of a derivative.

This inverse function of the derivative is called integration.

Here the given function is: g(t) = 8 + t² + t

Therefore, the antiderivative of the given function is

∫g(t) dt

= ∫(8 + t² + t) dt

= ∫8 dt + ∫t² dt + ∫t dt

= [8t⁽⁰⁺¹⁾/(0+1) + t⁽²⁺¹⁾/(2+1) + t⁽¹⁺¹⁾/(1+1) + c]

= (8t + t³/3 + t²/2 + c)

Here 'c' is the constant.

Again, differentiating the result, we get:

d/dt(8t + t³/3 + t²/2 + c)

= [8 ˣ 1 ˣ t⁽¹⁻¹⁾ + 3 ˣ t⁽³⁻¹⁾/3 + 2 ˣ t⁽²⁻¹⁾/2 + 0]

= 8 + t² + t

= g(t)

The antiderivative of the given function g(t)is (8t + t³/3 + t²/2 + c).

Learn more about antiderivative here: brainly.com/question/20565614

#SPJ4

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This task would be especially well-suited for instructional purposes. Students will benefit from a class discussion about the slope, y-intercept, x-intercept, and implications of the restricted domain for interpreting more precisely what the equation is modeling.

One potential confusion students may have follows from the subtle difference between what the car is doing and the idea of slope as the ratio between the change in vertical distance on the graph and the change in horizontal distance on the graph. Because the car is traveling one mile on a down-hill slope, the situation could be represented as a right triangle with a hypotenuse of 5,280 ft and a leg of 250 ft; using the Pythagorean Theorem they would find that the other leg is approximately 5,274 ft. Following through on this interpretation, a student might conclude that the car travels a horizontal distance of approximately 5,274 ft for every 250 ft in vertical distance and arrive at a slope of approximately -0.047. While this is, in some sense, the slope of the hill, it is not the slope of the function as described. This interpretation yields numbers that are very close to the situation described in the task, yet conceptually different since the distance traveled by the car would now be expressed in terms of horizontal distance traveled as opposed to distance along the slope of the hill to compute the elevation. If students do indeed pursue this line of reasoning, the task provides an opportunity to compare and contrast the graph of the function and what it represents with a drawing of the hill and the vertical and horizontal distances traversed with each mile down the slope.

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