Last calc question, I swear. This is my last question for the night. For the given function, find the points on the graph at which the tangent line has slope 1. y = 1/3x^3 - 1/2x^2 +x
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Originally Posted by aphan19 Last calc question, I swear. This is my last question for the night. For the given function, find the points on the graph at which the tangent line has slope 1. y = 1/3x^3 - 1/2x^2 +x Tangent lines slope at where f(x) is the curve the tangent line is tangent to...so find x such that
Originally Posted by aphan19 Last calc question, I swear. This is my last question for the night. For the given function, find the points on the graph at which the tangent line has slope 1. y = 1/3x^3 - 1/2x^2 +x Find y'. Set it equal to 1 to find the x values at which the derivative is equal to one: when or . Hope that answered your question!
Why is it when I take the derivative of 1/3x^3 its x^2? I thought I was suppose to move the x^3 to the top making it x^-3?
Originally Posted by aphan19 Why is it when I take the derivative of 1/3x^3 its x^2? I thought I was suppose to move the x^3 to the top making it x^-3? So did you mean or ?
Nvm, I got it. thank you very much though.
If you mean , the derivative would then be . To see where it's equal to 1: Multiply through by to clean this up: . Is this what you're looking for?
Originally Posted by aphan19 Nvm, I got it. thank you very much though. Ok. I ended up working out the other problem...you can ignore that...if you want
Originally Posted by Chris L T521 Ok. I ended up working out the other problem...you can ignore that...if you want You uhm...might want to wait until they respond to your question before you go ahead and asnwer it possibly confusing them
Originally Posted by Mathstud28 You uhm...might want to wait until they respond to your question before you go ahead and asnwer it possibly confusing them Yea. Thanks for that. I should have waited.
Originally Posted by Chris L T521 Yea. Thanks for that. I should have waited. Don't worry! I applaud your enthusiasm!
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