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L'Hospital's Rule, Asymptotes, Global Extrema, Inflection Points and Concavity : f(x) = x^2 e^(17x)

F(x) = x^2 e^(17x)

1. Find an equation for each horizontal asymptote to the graph of f.
2. Find an equation for each vertical asymptote to the graph of f.
3. Determine all critical numbers.
4. Determine the global maximum of the function.
5. Determine the global minimum of the function.
6. Find inflection points.
7. Find intervals of concave downward.
8. Give a rough sketch of your graph

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Hi, here is the solution,

f(x)= x^2 e^17x.

(i) Horizontal asymptote:

There are no asymptotes, because the curve is a simple polynomial.

Note: For simple polynomial asymptotes does not exist.

Critical point:

To find the critical point, we have to find the derivative and equate to zero, then find the values for x.

f(x) = x^2. e^17x

f '(x) = 2x. e^17x +17 x^2. e^17x

f'(x) = xe^17x( 2+ 17x)

Now, make f '(x)=0

so, x e^17x (2+17x) =0

2+17x = 0

17x = ...

Solution Summary

L'Hospital's Rule, Asymptotes, Global Extrema, Inflection Points and Concavity are investigated for f(x) = x^2 e^(17x). The solution is detailed and well presented.

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