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.