Let f(x) = ∫ [-2, x^2 - 3x] e^t^2 dt
At what value of x is f(x) a minimum?
If f(x) = ∫[a,u] g(t) dt then df/dx = g(u) * du/dx
This is just the chain rule in reverse
df/dx = e^(x^2-3x)^2 (2x-3)
f(x) is a min or max when x = -3/2
So now, is it a min or a max? How can you tell?
Well, finding the minimum of a function can sometimes be a daunting task. It's like searching for a parking spot in a crowded parking lot – you have to navigate through all those cars, be patient, and sometimes even get honked at! But fear not, because I have a solution that's sweeter than finding a parking spot right next to the exit.
To find the value of x for which f(x) is a minimum, we'll need to use some calculus wizardry. First, let's find the derivative of f(x) with respect to x.
Just like a magician pulling a rabbit out of a hat, we'll use the Fundamental Theorem of Calculus to make this differentiation much easier. The derivative of f(x) is simply the integrand evaluated at the upper limit times the derivative of the upper limit.
So, we have f'(x) = (e^(x^2 - 3x)^2) * (2x - 3).
Now, for f(x) to have a minimum, its derivative f'(x) must be equal to zero. So, we set 2x - 3 = 0 and solve for x.
Solving this equation would be like a puzzle - you have to move the pieces around until they match. After a bit of mathematical maneuvering, we find x = 3/2.
Voila! We've found the value of x for which f(x) is a minimum. It's as satisfying as finding the last piece of a puzzle and completing it. So, x = 3/2 is the answer you're looking for.
To find the value of x at which f(x) is a minimum, we need to find the critical points of the function f(x). The critical points are the values of x where the derivative of f(x) is equal to zero or does not exist.
Let's start by finding the derivative of f(x) with respect to x. We'll use the Fundamental Theorem of Calculus to differentiate the integral:
f'(x) = d/dx ∫ [-2, x^2 - 3x] e^t^2 dt
According to the Fundamental Theorem of Calculus, the derivative of an integral with respect to x is the integrand evaluated at the upper limit multiplied by the derivative of the upper limit. In this case, the upper limit is x^2 - 3x.
f'(x) = e^(x^2 - 3x)^2 * (2x - 3)
Now, we have the derivative of f(x). To find the critical points, we set f'(x) equal to zero and solve for x:
2x - 3 = 0
2x = 3
x = 3/2
The critical point of f(x) occurs at x = 3/2.
To determine if this critical point represents a minimum, maximum, or inflection point, we can analyze the second derivative of f(x). We'll take the derivative of f'(x):
f''(x) = d/dx (2x - 3)
f''(x) = 2
Since f''(x) is a constant value of 2, it does not change sign with x. This means that the critical point at x = 3/2 represents a minimum point of f(x).
Therefore, f(x) has a minimum at x = 3/2.
To find the value of x at which f(x) is a minimum, we need to find the critical points of f(x). Critical points are the values of x where the derivative of f(x) is equal to zero or undefined.
Let's start by finding the derivative of f(x) with respect to x. We can apply the Fundamental Theorem of Calculus to differentiate an integral with limits using the Chain Rule. The derivative of f(x) is given by:
f'(x) = d/dx [∫[-2, x^2 - 3x] e^t^2 dt]
To differentiate the integral with respect to x, we use the Fundamental Theorem of Calculus:
f'(x) = e^[(x^2 - 3x)^2] * d/dx[(x^2 - 3x)]
= e^[(x^2 - 3x)^2] * (2x - 3)
Now, to find the critical points, we set f'(x) equal to zero:
e^[(x^2 - 3x)^2] * (2x - 3) = 0
Now, we have two factors: e^[(x^2 - 3x)^2] = 0 or (2x - 3) = 0. However, e^[(x^2 - 3x)^2] cannot be zero since it is always positive. So we only need to solve (2x - 3) = 0:
2x - 3 = 0
2x = 3
x = 3/2
Therefore, the critical point is x = 3/2. To determine whether it is a minimum or maximum, we can apply the second derivative test:
To find the second derivative of f(x), we differentiate f'(x):
f''(x) = d/dx [e^[(x^2 - 3x)^2] * (2x - 3)]
= e^[(x^2 - 3x)^2] * (2x - 3)^2 + e^[(x^2 - 3x)^2] * 2
Evaluating f''(x) at x = 3/2:
f''(3/2) = e^[(3/2)^2 - 3(3/2))^2] * (2(3/2) - 3)^2 + e^[(3/2)^2 - 3(3/2))^2] * 2
After calculating f''(3/2), if the value is positive, then x = 3/2 is a minimum. If the value is negative, then x = 3/2 is a maximum.
Therefore, by following these steps, you can determine whether x = 3/2 is the value of x at which f(x) is a minimum.