# Show that the local maximum value of is less than local minimum value.

Let To find what are the local maximum/minimum values we first have to find where the function has its local minima/maxima and for that we need to find its critical points where f’(x) = 0

Differentiate with respect to x and equate to 0  x2 – 1 = 0

(x + 1)(x – 1) = 0

x = -1 and x = 1

Hence we have two critical points and we have to check first which point is local minima/maxima

Now to check whether x = 1 or x = -1 is a point of minima or maxima we have to check behaviour of f’’(x). If f’’(x) is negative at x = 1 or x = -1(that is f’’(1) or f’’(-1) < 0) then x = 1 or x = -1 is a point of maxima and else if f’’(x) is positive at x = 1 or x = -1(that is f’’(1) or f’’(-1) > 0) then x = 1 or x = -1 is a point of minima

We have calculated f’(x) Differentiate again with respect to x Observe that when we put x = 1 f’’(1) > 0 hence x = 1 is a point of minima and when we put x = -1 f’’(-1) < 0 hence x = -1 is a point of maxima

Now we have got where f(x) attains local maxima/minima, let us find what are the maximum/minimum values

Put x = 1 in f(x) to get local minima f(1) = 2

Hence local minimum value is 2

Put x = -1 in f(x) to get local maxima f(-1) = 0

Hence local maximum value is 0

Hence we conclude that the local maximum value (0) of f(x) is less than its local minimum value (2)

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