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?
To determine whether the decreasing arc of a function reaches a local or global minimum, we need to analyze the behavior of the function as it approaches negative infinity (or as x goes towards negative infinity).
To find the limit of f(x) as x approaches negative infinity, we can substitute negative infinity into the function and determine the behavior of the function.
a. Let's analyze the behavior of f(x) as x approaches negative infinity:
lim(x→-∞) e^2x(x^2-2)
As x approaches negative infinity, the exponential term e^2x approaches zero since the exponent becomes more and more negative. Similarly, the term (x^2 - 2) also approaches infinity since the square term dominates.
Therefore, the limit of f(x) as x approaches negative infinity is negative infinity.
This means that as x approaches negative infinity, the function f(x) decreases without bound towards negative infinity. However, since we are analyzing the behavior as x approaches negative infinity, we cannot determine if it reaches a local or global minimum.
b. To determine if the function f(x) has a global maximum, we need to analyze its behavior as x approaches positive infinity (or as x goes towards positive infinity).
Let's analyze the behavior of f(x) as x approaches positive infinity:
lim(x→∞) e^2x(x^2-2)
As x approaches positive infinity, both the exponential term e^2x and the term (x^2 - 2) approach infinity. Therefore, the limit of f(x) as x approaches positive infinity is positive infinity.
This means that as x approaches positive infinity, the function f(x) increases without bound towards positive infinity. Since the function increases without bound, it does not have a global maximum.
In conclusion:
a. The decreasing arc of the function f(x) = e^2x(x^2-2) does not reach a local or global minimum as x approaches negative infinity.
b. The function f(x) = e^2x(x^2-2) does not have a global maximum.