Question

Maxima , minima :

Answer 1

\[f(x) = \frac{ x^2 + 9 }{ x } \]

Answer 2

Find the upwards and downwards concavity and the turning points if they exists.

Answer 3

@Michele_Laino

Answer 4

we have to compute the first derivative of your function f(x)

Answer 5

I did , the whole problem is that the turining point is (0 , undefined ) which is senseless for me.

Answer 6

I got this:
\[\Large f'\left( x \right) = \frac{{2x \cdot x - \left( {{x^2} + 9} \right) \cdot 1}}{{{x^2}}} = \frac{{{x^2} - 9}}{{{x^2}}}\]

Answer 7

now we have to solve this inequality:
\[\frac{{{x^2} - 9}}{{{x^2}}} \geqslant 0\]

Answer 8

since at those points at which f' is positive, the function f is increasing

Answer 9

what is the solution of the inequality above?

Answer 10

Concavity not maxima ,

Answer 11

yes! I know it

Answer 12

x^2 - 9 > x^2
-9 > 0

Answer 13

the first derivative is positive, when the numerator only is positive, since the denominator is always positive in all points x such that x is different from zero

Answer 14

so we have to solve this equivalent inequality:
\[{x^2} - 9 \geqslant 0\]

Answer 15

I got laggy and disconnected.

Answer 16

3 > x > -3

Answer 17

the solution is:
\[x \geqslant 3,\quad x \leqslant - 3\]
so, we can plot your solution as below: