# Thread: Help in the quick way to diff y=5-X^2

1. ## Help in the quick way to diff y=5-X^2

I need vertex,
line of Sym x=
Max and Min
is the F(0)=5 a min or a max?

Much appriciated

2. vertex is (0,5)

line of sym is x=0

take derative of function and equate to 0.

y'= -2x

0= -2x

so x = 0 this is either a min or max. Now you can use the 2nd derative test or plug in values for the 1st derative and see where it increasing or decreasing. For example -1<0<1

y' = -2(-1)= 2
y' = -2(1) = -2

so when x=0 is a max

Hope that helps

3. You don't need to differentiate.

We are looking for the maximum value of $5 - x^2$. It's obvious that $x^2$ is always positive or zero. So, the minimum of $x^2$ is 0. Therefore $5-x^2$ is maximum at $x=0$.

We can also find it using differentials. The local extrema of a function are at the roots of $y' = 0$. Don't forget to check for global extrema too.

$y' = -2x = 0$

$x=0$

So we have a local extremum at $x=0$.

We can decide whether it's maximum or minimum using the second derivative test.

$y''(0) = -2 < 0$. It's a maximum.

Don't forget to check for global maximum too. The reason for doing this is, the function may have a maximum at infinity.

$\lim_{x\to\infty} 5-x^2 = -\infty$
$\lim_{x\to-\infty} 5-x^2 = -\infty$

Ok, now we're convinced that the function has a maximum only at $x=0$.