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have you studied calculus?
yes. i think i should take the first derivative of this function and minimize it on the interval \[\left( -\infty ,\infty \right)\] i found that at x=4/3, it's slope will be least. i don't know what to do next.
well the equation of the tangent at x = 4/3 is value of the the function at x=4/3 just plug in this value of into f(x)
differentiate 3x^2 - 8x + 5 = 0 for critical points (3x -5)(x - 1) = 0 x = 5/3 or x = 1 there are two turning points second derivative = 6x-8 so x = 5/3 gives minimum and x=1 gives maximum
there are two tangents with slope zero at x=1 and at x=5/3
ah. but when you get the critical points of the first derivative, you are minimizing the value of the original function and not its slope. am i making sense?
yes the critical points give you local minimum and local maximum of the original function or it could be a point of inflection. The tangents to the graph at these max and min points will have 0 slope. The question is a bit confusing it asks for the tangent (singular) with least slope but there are two of these. i'll try and draw it
excuse the bad drawing!
i think we can get a single tangent line by just minimizing the second first derivative. love the drawing.