Originally posted by Stright Success
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[HELP!!!] Calc test!!!
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Product rule: f(x)*g'(x)+ f '(x) * g(x)
f(x) = x * e^x
f'(x) = x *e^x + e^x
And I don't believe it has any inflections points (points at which the graph changes concavity, i.e. concave up to concave down). Use your graphing calculator!Last edited by d2privat11; 02-06-2007, 11:49 PM.
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My bad (I sometimes rush) but I still get A's.
Here ya go:
Chain Rule = f'(x)(g(x))g'(x)
Multi. Rule = f'(x)g(x)+f(x)g'(x)
Quot. Rule = f'(x)g(x)-f(x)g'(x)/g(x)^2
Inflection Points are found by finding a (c) critical number.
Then evaluate your entire domain at your (c)'s to find where the the graph's derivative shifts from positive to negative or vice versa. That's an inflection point. I think, this is right. Haven't used it in a while. This should be right.The Lord watches over me!
"Stop punching down on my people!!!"
- D. Chappelle
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Originally posted by d2privat11Product rule: f(x)*g'(x)+ f '(x) * g(x)
f(x) = x * e^x
f'(x) = x *e^x + e^x
And I don't believe it has any inflections points (points at which the graph changes concavity, i.e. concave up to concave down). Use your graphing calculator!
df/dx = x * e^x + e^x
Yea I know he put it, this is just a confirmation. My bad for the mishap.The Lord watches over me!
"Stop punching down on my people!!!"
- D. Chappelle
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http://integrals.wolfram.com/index.jsp
I lived off that in college math.
Searching Google is key...and that's why I'm a semester away from being an engimaneer, it's my research skills.
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Originally posted by AccordWarriorhttp://integrals.wolfram.com/index.jsp
I lived off that in college math.
Searching Google is key...and that's why I'm a semester away from being an engimaneer, it's my research skills.
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