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If f(x)= ∫ (16−t^2)(e^(t^5))dt , on what interval is f increasing?
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Answer (in interval notation):
To determine on what interval the function f(x) is increasing, we need to find the intervals where the derivative of f(x) is positive.
First, let's find the derivative of f(x) using the Fundamental Theorem of Calculus. Since the upper limit of the integral is x, we can consider f(x) as a function of x.
Let F(x) be the antiderivative of (16 - t^2)(e^(t^5)) with respect to t:
F(x) = ∫ (16 - t^2)(e^(t^5))dt
Now, we can find the derivative of f(x) by differentiating F(x) with respect to x:
f'(x) = d/dx [∫ (16 - t^2)(e^(t^5))dt]
Since F(x) is an indefinite integral, to differentiate it with respect to x, we need to apply the Chain Rule:
f'(x) = [dF(x)/dx] = [(dF(x)/dt) * (dt/dx)].
To find (dF(x)/dt), we differentiate F(x) with respect to t:
f'(x) = (d/dt)[∫ (16 - t^2)(e^(t^5))dt] * (dt/dx)
Note that (dt/dx) is 1 since t is the integration variable.
Simplifying, we have:
f'(x) = ∫ (16 - t^2)(e^(t^5))dt * (dt/dx)
Now, we can simplify the expression (16 - t^2)(e^(t^5)) * (dt/dx) and see if it is positive on any interval.
After simplifying and substituting dt for dx, the expression becomes:
f'(x) = (16 - t^2)(e^(t^5))
To find the intervals where f'(x) > 0, we need to solve the inequality:
(16 - t^2)(e^(t^5)) > 0
To solve this inequality, we need to consider the sign of each factor:
For (16 - t^2) to be positive, we have:
1) 16 - t^2 > 0
t^2 < 16
-4 < t < 4
For (e^(t^5)) to be positive, we have:
2) e^(t^5) > 0
This is true for all values of t.
Now, we combine the conditions (1) and (2) to find the intersection:
-4 < t < 4
Therefore, the function f(x) is increasing on the interval (-4, 4).
In interval notation:
(-4, 4)
you know that f' = (16-x^2) e^(x^5)
where is that positive?