Question

Find the equation of the line tangent to the graph of x^3 - 4x^2 + 5x with the least slope. What can I do on this one?

Answer 1

have you studied calculus?

Answer 2

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.

Answer 3

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)

Answer 4

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

Answer 5

there are two tangents with slope zero at x=1 and at x=5/3

Answer 6

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?

Answer 7

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