Given the function f(x)= e^2x(x^2-2)
a. does the decreasing arc reach a local or global minimum?
b. does f have a global max?
f = e^2x(x^2-2)
f ' = 2e^2x(x-1)(x+2)
f '' = 2e^2x(2x^2+4x-3)
f'=0 at x=-1,2
f''(-2) < 0, so a max at x = -1
f''(1) > 0, so a min at x=2
Since f(x) --> 0 as x --> -oo, the minimum is a global min.
since f(x) --> +oo as x --> +oo, there is no global max
a. Well, let's see. The function f(x) = e^2x(x^2-2) seems to have a product of two factors: e^2x and (x^2-2). The exponential factor e^2x is always positive, so it won't affect whether the function reaches a minimum or not. As for the second factor, (x^2-2), it is actually a parabola that opens upwards (because the coefficient of x^2 is positive). So, if we look at the behavior of (x^2-2), it reaches a local minimum point when x = 0. Therefore, the decreasing arc of f(x) will reach a local minimum.
b. Now, in terms of a global maximum, we need to consider the entire function f(x) = e^2x(x^2-2). As x approaches positive infinity or negative infinity, e^2x grows without bound, which means f(x) also grows without bound. So, f(x) does not have a global maximum. It's like a never-ending race to infinity!
To determine whether the given function f(x) = e^2x(x^2 - 2) reaches a local or global minimum and if it has a global maximum, we need to find the critical points and determine the behavior of the function around those points.
a. To find the critical points, we take the derivative of f(x) with respect to x:
f'(x) = 2e^2x(x^2 - 2) + e^2x(2x)
= 2e^2x(x^2 - 2 + 2x)
= 2e^2x(x^2 + 2x - 2)
Setting f'(x) equal to zero and solving for x, we have:
2e^2x(x^2 + 2x - 2) = 0
From this equation, we can see that there are three factors multiplying to zero. One of them must be equal to zero.
The first factor, 2e^2x, is never equal to zero.
The second factor, (x^2 + 2x - 2), can be solved using the quadratic formula to find its roots:
x = (-2 ± √(2^2 - 4*1*(-2))) / (2*1)
x = (-2 ± √(4 + 8)) / 2
x = (-2 ± √12) / 2
x = (-2 ± 2√3) / 2
x = -1 ± √3
Therefore, the second factor has two critical points: x = -1 + √3 and x = -1 - √3.
b. To determine if f(x) has a global maximum, we need to analyze the behavior of the function for x approaching positive or negative infinity.
As x approaches positive or negative infinity, the term e^2x becomes exponentially large, dominating all other terms in f(x). Hence, f(x) goes to positive infinity as x approaches positive or negative infinity.
Since f(x) goes to positive infinity and doesn't have any critical points where f'(x) is equal to zero, we conclude that f(x) does not have a global maximum.
In summary:
a. The decreasing arc of f(x) reaches two local minima at x = -1 + √3 and x = -1 - √3.
b. f(x) does not have a global maximum.