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Let f(x) = integral from (-2) to (x^2-3x) of e^(t^2) dt. At what value of x is f(x) a minimum?

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2 Answers

Let f(x) = integral from (-2) to (x^2-3x) of e^(t^2) dt. At what value of x is f(x) a minimum?
 
Notice, because of our excellent continuity, we have:
 
dt/dx = 2x-3
 
and
 
d/dx = d/dt * dt/dx
 
and we have:
 
df/dt = e^(t^2)
 
therefore:
 
df/dx = d(∫(-2)(x^2-3x) df/dt dt ) / dx
 
= {d[∫(df/ dt ) dt] / dt |(-2)(x^2-3x) }* dt/dx
 
But the derivative of the integral by the same variable is the original thing, as long as we are careful accounting for taking the derivative of the constant term t = -2. This is just:
 
= {(df/ dt )|(x^2-3x) }* dt/dx
 
substituting and evaluating we get:
 
= {(e^(t^2) )|t= (x^2-3x) } * (2x-3)
 
= e^((x^2-3x)^2) * (2x-3)
 
this has a critical point precisely when x = 3/2. If x < 3/2 this expression is negative. If x > 3/2 this expression is positive, so we have a minimum.
 
Sometimes doing the computations just gets in the way.
 
 
f(x) is at a minimum if f'(x) = 0 and  f"(x) > 0. This is because, at a minimum, the original function must have zero slope and be concave up (like a cup). For a maximum, the original function would again have zero slope, but would be concave down (like a frown).
 
However, e^((x^2-3x)^2) = 0 has no solution. Therefore, there can be no minimum or maximum.
 
For practice, I'll find f"(x) using the Chain Rule:
 
f"(x) = e^((x^2 - 3x)^2) * 2(x^2 - 3x) * (2x - 3)

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