I said f'(x) = m = 0 = 2xe^2x ---> x = 0/(2xe^2x) = 0 Is this right?
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Originally Posted by satx I said f'(x) = m = 0 = 2xe^2x ---> x = 0/(2xe^2x) = 0 Is this right? Your method is right although you need to use the product rule when finding
Wouldn't that equal 2?
Originally Posted by satx Wouldn't that equal 2? i believe you have,
to solve
Ok then that would equal one then, if x = 0, no? (e^2(0))(0) + e^2(0)) = 0 + 1 = 1???
Originally Posted by Raoh to solve So e^0 is 0? I'm so confused
Originally Posted by satx Ok then that would equal one then, if x = 0, no? (e^2(0))(0) + e^2(0)) = 0 + 1 = 1??? i think you're trying to solve
Originally Posted by Raoh i think you're trying to solve Well isn't the slope equal to the derivative?
Originally Posted by satx So e^0 is 0? I'm so confused well ,what i wrote means and the you're searching for is .
Originally Posted by satx Well isn't the slope equal to the derivative? yes
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