Question

How do you find the local max and min for `f(x) = 2x^3 - 5x +1` in the interval is (-3,3)?

Answer 1

We have a local minima at `x=+sqrt(5/6)` and a local maxima at `x=-sqrt(5/6)`.

Explanation:

A maxima is a high point to which a function rises and then falls again. As such the slope of the tangent or the value of derivative at that point will be zero.

Further, as the tangents to the left of maxima will be sloping upwards, then flattening and then sloping downwards, slope of the tangent will be continuously decreasing, i.e. the value of second derivative would be negative.

A minima on the other hand is a low point to which a function falls and then rises again. As such the tangent or the value of derivative at minima too will be zero.

But, as the tangents to the left of minima will be sloping downwards, then flattening and then sloping upwards, slope of the tangent will be continuously increasing or the value of second derivative would be positive.

However, these maxima and minima may either be universal i.e. maxima or minima for the entire range or may be localized, i.e. maxima or minima in a limited range.

Let us see this with reference to the function described in the question and for this let us first differentiate `f(x)=2x^3-5x+1`.

Its first derivative is given by `f'(x)=6x^2-5` which would be zero for `6x^2-5=0` or `x=+-sqrt(5/6)`. These are within the range `(-3,3)`.

Hence maxima or minima occurs at points `x=+sqrt(5/6)` and `x=-sqrt(5/6)`.

To find whether it is maxima or minima, let us look at second differential which is `f''(x)=12x` and hence while

at `x=+sqrt(5/6)`, `f''(x)=12sqrt(5/6)` and is positive

at `x=-sqrt(5/6)`, `f''(x)=-12sqrt(5/6)` and is negative

Hence, we have a local minima at `x=+sqrt(5/6)` and a local maxima at `x=-sqrt(5/6)`.

Please see the graph below, it is apparent that function does take values less than minima and more than maxima i.e. these are local maxima and minima.

graph{2x^3-5x+1 [-10, 10, -5, 5]}