Asked by Suzy
22. Divide the expression before differentiating.
F(x) = (8x^3 - 1)/ (2x - 1)
Would the way to solve this problem be...
F(x) = [(8x^3) / (2x - 1)] - [(1)/(2x - 1)]
= [(4x^2)/(1)] - [(1) / (2x - 1)]
= (4x^2) - (2x-1)^-1
F'(x) = (8x + 1)
F(x) = (8x^3 - 1)/ (2x - 1)
Would the way to solve this problem be...
F(x) = [(8x^3) / (2x - 1)] - [(1)/(2x - 1)]
= [(4x^2)/(1)] - [(1) / (2x - 1)]
= (4x^2) - (2x-1)^-1
F'(x) = (8x + 1)
Answers
Answered by
bobpursley
No.
The numerator is a difference of two squares, factor it, then, the denominator divides out evenly. Your answer is wrong.
The numerator is a difference of two squares, factor it, then, the denominator divides out evenly. Your answer is wrong.
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