Finding minima of $x^2$ function with derivatives

Question

Of course it is $0$.

with $-8$ ... $f(x^2): 64$

The derivative would be $f(2x): -16$

The derivative tells me about change and slope. From $x=-8$ and calculating derivatives I know I need to go down because of the negative sign. But I really don't understand what to do next? What does $-16$ tell me at all? To move $-16$ on $x$ axis? This way I get to: $-8-16 = -24$

$f(x^2): 576$. So it is obvious that this is wrong.

If I go $-16$ on $y$ axis I get $48$. This would make sense, but this means root of $48$ is my $x$.

But guys, how should I calculate the root of $48$ with no calculator? And function is so simple...

So I know I am missing something, I would appreciate if you could help me. Thank you.

Answer 1

For a differentiable function as $f(x)=x^2$, a necessary condition for a max/min point is that $f’(x)=2x=0\implies x=0$.

Note that it is not sufficient to determine what kind of point is x=0, we need to consider the sign of $f’(x)$ or to consider $f’’(x)$.

In this case

• for $x>0 \implies f’(x)>0$ that is $f$ increasing

• for $x<0 \implies f’(x)<0$ that is $f$ decreasing

thus $x=0$ is a minimum point.

Answer 2

Here, you can construct a variation table, for instance:

x |  ...  | x1 |  ...  | x2 |  ...
y'|  +/-  | 0  |  +/-  | 0  |  +/-
y |  +/-  | y1 |  +/-  | y2 |  +/-

If $y'=f'(x)=\text{negative }(-)$, this means that the rate of change is negative, and that $f(x)$ at that point is increasing.