stationary point
Determine the stationary points of the following function and for each stationary point determine
whether it is a local maximum, local minimum or a point of inflexion.
$f(x)=x^3(x-1)^2$
Using product rule to find $dy/dx$ I got:
$5x^4-8x^3+3x^2$ how can I factorize this and get the two $x$ values?
$5x^4-8x^3+3x^2=0$ how to find the two values from here?
Take the second derivatives to determine whether a function is maximized or minimized at a point.
$4^+$ means approaching the point $4$ from above.
Hmm. \begin{align*} y&=\frac{1}{x}+\frac{1}{x^2}-\frac{1}{x^3} \\ &=\frac{x^2+x-1}{x^3} \\ y'&=\frac{x^3(2x+1)-3x^2(x^2+x-1)}{x^6} \\ &=\frac{x(2x+1)-3(x^2+x-1)}{x^4} \\ &=\frac{2x^2+x-3x^2-3x+3}{x^4} \\ &=\frac{-x^2-2x+3}{x^4}. \end{align*} Setting this equal to zero is tantamount to solving $x^2+2x-3=0,$ with solutions $$x=\frac{-2\pm\sqrt{4+4(3)}}{2}=\frac{-2\pm 4}{2}=\{1, -3\}. $$ You can probably use the Second Derivative Test to show which of these is a local min, and which a local max.
Best Answer
Notice that, by factoring out $x^2$,
$$5x^4 - 8x^3 + 3x^2 = x^2(5x^2 - 8x + 3) = 0$$
Obviously, the left factor in the result gives you $x=0$ as a potential zero of $f'(x)$. You can use your method of choice, e.g. the quadratic formula, to find the zeroes of the quadratic factor and obtain your other zeroes for $f'(x)$.