For the function f(x)=x^3e^x, determine intervals of increase and decrease and the absolute minimum value of f(x)
To do this I found the derivative which is f'(x)=3x^2e^x.
Then I set it to zero:
f'(x)=0
0=3x^2e^x
How would I isolate x? And when I do, I would sub that x value into the original function to find the intervals of increase and decrease right? Also, how would I find the absolute minimum value?
I made a mistake... i figured it out
To isolate x in the equation 0=3x^2e^x, you can start by dividing both sides of the equation by 3x^2:
0 = e^x
Next, take the natural logarithm (ln) of both sides of the equation:
ln(0) = ln(e^x)
Since the natural logarithm of 0 is undefined, we can conclude that there is no real value of x that satisfies the equation 0=3x^2e^x. Therefore, there is no critical point or solution in this case.
To determine the intervals of increase and decrease for the function f(x)=x^3e^x, you can make use of the first derivative test. Since there are no critical points, the function does not have any intervals of increase or decrease.
To find the absolute minimum value of f(x), you need to look for any local minimum points on the graph. Since there are no intervals of increase or decrease, the function does not have a local minimum value. Therefore, it does not have an absolute minimum value either.
In summary, the function f(x)=x^3e^x does not have any intervals of increase or decrease, nor does it have any absolute minimum value.