There are no inflection points on an exponential graph, e.g. ask wolfram alpha "inflection exp". The curvature of an exponential is always the same sign, in this example it is curved up or positive curvature. Notice the second derivative of y=e^(bx) is y''=b^2 e^(bx)=b^2 y which is always positive. Since your equation x=(ln(1/b))/b finds the x, for which y=1/b, you are actually finding the point that has a slope of 1, since y'=by = b (1/b) = 1.
There are no inflection points on an exponential graph, e.g. ask wolfram alpha "inflection exp". The curvature of an exponential is always the same sign, in this example it is curved up or positive curvature. Notice the second derivative of y=e^(bx) is y''=b^2 e^(bx)=b^2 y which is always positive. Since your equation x=(ln(1/b))/b finds the x, for which y=1/b, you are actually finding the point that has a slope of 1, since y'=by = b (1/b) = 1.
thehouseofbrawn 2 months ago