Given that y = 8x + 2x^-1, find the 2 values for x for which dy/dx = 0
First differentiate y with respect to x, which gives you dy/dx = 8 - 2x^-2. This needs to equal zero so equate to zero. 8-2x^-2 = 0. You can then bring the 2x^-2 to the other side giving 2x^-2=8. Dividing both sides by 2 gives x^-2 = 4. You can then flip both sides, giving x^2 = 1/4. Then square root both sides giving x = +/- 1/2.
RB
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