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Math  /  Calculus

QuestionIf f(x)=∫0x(16−t2)et3dtf(x)=\int_{0}^{x}\left(16-t^{2}\right) e^{t^{3}} d t for all xx, then find the largest open interval on which ff is increasing. Answer (in interval notation): □\square

Studdy Solution

STEP 1

1. The function f(x)=∫0x(16−t2)et3 dt f(x) = \int_{0}^{x} (16 - t^2) e^{t^3} \, dt is defined as an integral from 0 to x x .
2. To determine where f(x) f(x) is increasing, we need to find where its derivative f′(x) f'(x) is positive.
3. The Fundamental Theorem of Calculus will be used to find the derivative of the integral.

STEP 2

1. Apply the Fundamental Theorem of Calculus to find f′(x) f'(x) .
2. Determine the sign of f′(x) f'(x) to find where it is positive.
3. Identify the largest open interval where f′(x)>0 f'(x) > 0 .

STEP 3

Apply the Fundamental Theorem of Calculus to find f′(x) f'(x) :
According to the Fundamental Theorem of Calculus, if F(x)=∫axg(t) dt F(x) = \int_{a}^{x} g(t) \, dt , then F′(x)=g(x) F'(x) = g(x) .
Therefore, for f(x)=∫0x(16−t2)et3 dt f(x) = \int_{0}^{x} (16 - t^2) e^{t^3} \, dt , we have:
f′(x)=(16−x2)ex3 f'(x) = (16 - x^2) e^{x^3}

STEP 4

Determine the sign of f′(x)=(16−x2)ex3 f'(x) = (16 - x^2) e^{x^3} .
Since ex3>0 e^{x^3} > 0 for all real x x , the sign of f′(x) f'(x) depends solely on 16−x2 16 - x^2 .
16−x2>0 16 - x^2 > 0
Solve for x x :
16>x2 16 > x^2 −4<x<4 -4 < x < 4

STEP 5

Identify the largest open interval where f′(x)>0 f'(x) > 0 .
From the inequality −4<x<4 -4 < x < 4 , the largest open interval where f(x) f(x) is increasing is:
(−4,4) (-4, 4)
The largest open interval on which f f is increasing is (−4,4) \boxed{(-4, 4)} .

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