SOLUTION: I truly see why people say it's not the calculus that trips people up, it's the algebra....
Let f(x) = x^(5/3) + 5x^(2/3)
f'(x) = (5/3)x^(2/3) + (10/3)x^(-1/3) = 0
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-> SOLUTION: I truly see why people say it's not the calculus that trips people up, it's the algebra....
Let f(x) = x^(5/3) + 5x^(2/3)
f'(x) = (5/3)x^(2/3) + (10/3)x^(-1/3) = 0
someone
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Question 1001511: I truly see why people say it's not the calculus that trips people up, it's the algebra....
Let f(x) = x^(5/3) + 5x^(2/3)
f'(x) = (5/3)x^(2/3) + (10/3)x^(-1/3) = 0
someone told me you can factor out an x^(1/3) but how does that work when there is 5/3x and 10/3 there and ones an exponent of (2/3) and a negative exponent of 1/3...I am really thrown off by this.
Please explain
Thank you Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Let f(x) = x^(5/3) + 5x^(2/3)
f'(x) = (5/3)x^(2/3) + (10/3)x^(-1/3) = 0
someone told me you can factor out an x^(1/3) but how does that work when there is 5/3x and 10/3 there and ones an exponent of (2/3) and a negative exponent of 1/3...I am really thrown off by this.
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f'(x) = (5/3)x^(2/3) + (10/3)x^(-1/3) = 0
Since it's = 0, you can multiply by x^(1/3)
--> (5/3)x + 10/3 = 0
--> x = -2
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You didn't give a reason for f' being = 0, tho.
