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determine the equation of the tangent line to f(x)=15+8ln(x) at x=1
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\[\huge f(x)=15+8lnx \] \[\huge f'(x)=8[lnx]' = \frac{8}{x}\]
so at x=1, f'(1) = 8/1 = 8.. this is slope of the tangent line... f(1) = 15 + 8*ln1 = 15 + 8*0 = 15 so the graph goes through (1, 15) now that you have a point and the slope of the tangent line, put it in point-slope form to get equation of the tangent line.
point-slope form: \[\huge y-y_0=m(x-x_0) \] \[\huge y-(15)=8(x-(1)) \] \[\huge y-15=8(x-1) \] this is the equation of your tangent line...
thank you very much your a great help
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