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?

Question

At what value of x is f(x) a minimum?

 

I know f'(x) = e^((x^2-3x)^2) , but I don't know how to continue.

 

Thanks!

Answer 1

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.

 

 

Answer 2

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)