Find the points of local maxima, local minima and the points of inflection of the function $f (x) = x^5 - 5x^4 + 5x^3 -1$. Also find the corresponding local maximum and local minimum values.

Given: function $f (x) = x^5 - 5x^4 + 5x^3 -1$

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 $f (x) = x^5 - 5x^4 + 5x^3 -1$

Calculating the first derivative of f(x), i.e.,

$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\mathrm{x}^{5}-5 \mathrm{x}^{4}+5 \mathrm{x}^{3}-1\right)}{\mathrm{d} \mathrm{x}} \\ \text{When the sum rule of differentiation is applied followed by taking out all the constant term, we get} \\\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\mathrm{x}^{5}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(5 \mathrm{x}^{4}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(5 \mathrm{x}^{3}\right)}{\mathrm{dx}}-\frac{\mathrm{d}(1)}{\mathrm{dx}}$

$f' (x) = 5x\textsuperscript{4}-5.4x\textsuperscript{3}+5.3x\textsuperscript{2}-0$

$\Rightarrow f' (x) = 5x\textsuperscript{4}-20x\textsuperscript{3}+15x\textsuperscript{2}$

Equating the first derivative with 0 to find out the critical point,

$\\ {f'(x) = 0}\\ { \Rightarrow 5x\textsuperscript{4}-20x\textsuperscript{3}+15x\textsuperscript{2} = 0}\\ { \Rightarrow 5x\textsuperscript{2}(x\textsuperscript{2}-4x+3) = 0}\\ { \Rightarrow 5x\textsuperscript{2}(x\textsuperscript{2}-4x+3) = 0}\\$

Then splitting the middle term, we get $\Rightarrow 5x\textsuperscript{2}(x^2-3x-x+3) = 0$

$\\{ \Rightarrow 5x\textsuperscript{2}[ x(x-3)-1(x-3)] = 0}\\ { \Rightarrow 5x\textsuperscript{2}(x-1)(x-3) = 0}\\ \Rightarrow 5x\textsuperscript{2} = 0 \ or\ (x-1) = 0 \ or\ (x-3) = 0\\ { \Rightarrow x = 0 \ or\ x = 1 \ or\ x = 3\\$

Now we will find the corresponding y value by putting the numerous values of x in given function

$f (x) = x^5 - 5x^4 + 5x^3 -1$

$When \ x = 0, f(0) = 0\textsuperscript{5} - 5(0)\textsuperscript{4} + 5(0)\textsuperscript{3} - 1 = -1$

Hence the point is (0,-1)

$When \ x = 1, f(1) = 1\textsuperscript{5} - 5(1)\textsuperscript{4} + 5(1)\textsuperscript{3} - 1 = 1-5+5-1 = 0$

Hence the point is (1,0)

$When \ x = 3, f(3) = 3\textsuperscript{5} - 5(3)\textsuperscript{4} + 5(3)\textsuperscript{3} - 1 = 243-405+135-1 = -28\\$

Hence the point is (3,-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 maxima.

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