Q. 26

# Find the points of local maxima, local minima and the points of inflection of the function f (x) = x5 – 5x4 + 5x3 – 1. Also find the corresponding local maximum and local minimum values.

Given: function f (x) = x5 – 5x4 + 5x3 – 1

To find: the points of local maxima, local minima and the points of inflection of f(x) and also to find the corresponding local maximum and local minimum values.

Explanation: given f (x) = x5 – 5x4 + 5x3 – 1

We will find the first derivative of f(x), i.e.,

Applying the sum rule of differentiation and taking out the constant term, we get

f' (x) = 5x4-5.4x3+5.3x2-0

f' (x) = 5x4-20x3+15x2

Now to find the critical point we need to equate the first derivative to 0, i.e.,

f'(x) = 0

5x4-20x3+15x2 = 0

5x2(x2-4x+3) = 0

5x2(x2-4x+3) = 0

Now splitting the middle term, we get

5x2(x2-3x-x+3) = 0

5x2[ x(x-3)-1(x-3)] = 0

5x2(x-1)(x-3) = 0

5x2 = 0 or (x-1) = 0 or (x-3) = 0

x = 0 or x = 1 or x = 3

Now we will find the corresponding y value by substituting the different values of x in given function f(x) = x5 – 5x4 + 5x3 – 1,

When x = 0, f(0) = 05 – 5(0)4 + 5(0)3 – 1 = -1

Hence the point is (0,-1)

When x = 1, f(1) = 15 – 5(1)4 + 5(1)3 – 1 = 1-5+5-1 = 0

Hence the point is (0,0)

When x = 3, f(3) = 35 – 5(3)4 + 5(3)3 – 1 = 243-405+135-1 = -28

Hence the point is (0,-28)

Therefore, we see that

At x = 3, y has minimum value = -28. Hence x = 3 is point of local minima.

At x = 1, y has maximum value = 0. Hence x = 1 is point of local minima.

And at x = 0, y has neither maximum nor minimum value, hence this is point of inflection.

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