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f(x)= (1/2x)(ln(x^20)), (-1,0)…..find the equation of the tangent line. y=
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\[m _{t}=(f'(x)=0)\]
???
what class are you in
calculus
so i take the derivative of the given f(x) right ?
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yesss
so i found the equation for the first given function f(x) = i/2x and got y= -1/2x-1/2. is that correct? can someone please confirm this
sorry i meant the first function to be 1/2x not i/2x
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okay so first: f(x)=1/2x the derivative of this is f'(x)= -1/2x^2 then i plugged in for x, F'(-1)= -1/2(-1)^2 and got -1/2 as my slope. then using the formula y-y1 = m(x-x1) I plugged in for m and finally got the equation y= -1/2x-1/2
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