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Calculus/Calculus How derivatives affect the shape of a graph

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Expert: - 3/23/2009


Question
Will you please help me figure out these things I cant figure them out.  Thank you so much.

given the equation f(x)= 9*Sqrt(x)*e^(-x)

a) Find the interval on which f is increasing.
b) Find the interval on which f is decreasing.
c) Find the local maximum value of f.
D) Find the inflection point.
e) Find the interval on which f is concave up.
f) Find the interval on which f is concave down.


Answer

The Function
The function given is f(x) = (9√x)e^(-x).

The derivative of this function is done by use of the product rule.

We get (9√x)d(e^(-x))dx + d(9√(x)dx*e^(-x).

We know that d(e^(-x))dx = -e^(-x).
√x = x^0.5, so the derivative of 9√x is the derivative of 9x^0.5.
This is 4.5x^(-0.5) = 4.5/√x.

Putting both of these derivatives back into the equation gives us
(-9√x)e^(-x) + (4.5/x)e^(-x).  This can be rewritten as
-(9√x)e^(-x)((1 - 1/(2x)).  Exponentials are never 0.  
The only place that this could be 0 would be where 1 = 1/(2x).

Before solving this, not that at x=0, the derivative is underfined.

Back to solving, multiply both sides by 2x, giving 2x=1, or x=1/2.
So the intervals to look at are (-infinity,0), we can say that the function is undefined since there is a √ it the function.  The √ can't be taken when the value is less than 0.  The two intervals to look at are (0,1/2), and (1/2,infinity).  The value 1/4 is in the first interval and 1 is in the second interval.  Put these value into f'(x).  When f'(x) > 0, it's increasing.  Where f'(x) is less than 0, it's decreasing.  

It can be seen that f'(x) is positive for x < 1/2 and negatvie for
x > 1/2.  I have attached a veiw of the function.

The only inflection point would be where the second derivative was 0.
We know that f'(x) was -9e^(-x)((√x) - 1/(2√x)), so f"(x) can be found as a product rule between the function -9e^(-x) and
√x - 1 /(2√x).  The derivative of -9e^(-x) is 9e^(-x) and the derivate of √x - 1/(2√x) is 1/(x√x) + 1/(4(√x^3)).

To find the inflection point, set this to zero and check the solution against the graph to make sure it looks right.

We can see that f(x) is concave down from 0 out to the inflection point and f(x) is concave up from the inflection point out to infinity.

Scotto

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