# Question #d530f

Sep 20, 2015

See the explanation.

#### Explanation:

Lets write function in the form:

$f \left(x\right) = 4 - {x}^{2} , x \in \left[- 2 , 2\right]$
$f \left(x\right) = - 4 + {x}^{2} , x \in \left(- \infty , - 2\right) \cup \left(+ 2 , + \infty\right)$

1st derivative:

$f ' \left(x\right) = - 2 x , x \in \left[- 2 , 2\right]$
$f ' \left(x\right) = 2 x , x \in \left(- \infty , - 2\right) \cup \left(+ 2 , + \infty\right)$

$f ' \left(x\right) = 0$
$- 2 x = 0 \iff x = 0 , x \in \left[- 2 , 2\right] , x$ is critical point
$2 x = 0 \iff x = 0 , x \notin \left(- \infty , - 2\right) \cup \left(+ 2 , + \infty\right) , x$ is not critical point

Sign of 1st derivative:
$\forall x \in \left(- \infty , - 2\right) f ' \left(x\right) = 2 x < 0 , f$ is decreasing
$\forall x \in \left(- 2 , 0\right) f ' \left(x\right) = - 2 x > 0 , f$ is increasing
$\forall x \in \left(0 , + 2\right) f ' \left(x\right) = - 2 x < 0 , f$ is decreasing
$\forall x \in \left(+ 2 , + \infty\right) f ' \left(x\right) = 2 x > 0 , f$ is increasing

$f ' \left(x\right) = 2 x$ on the interval $\left[- 2 , 2\right]$ is continuous function, $f ' \left(x\right)$ changes sign in $x = 0$ hence function $f \left(x\right)$ has maximum value in $x = 0$ and ${f}_{\max} = f \left(0\right) = 4$.

In points $x = - 2$ and $x = 2$ may exist discontinuities of the 1st order, so we have to check if the function $f \left(x\right)$ is continuous in these points.

$f \left(- 2\right) = 0$
${\lim}_{x \to - {2}_{-}} f \left(x\right) = {\lim}_{x \to - {2}_{-}} \left(- 4 + {x}^{2}\right) = A$
$x = - 2 - \epsilon$
$A = {\lim}_{\epsilon \to 0} \left(- 4 + {\left(- 2 - \epsilon\right)}^{2}\right) = {\lim}_{\epsilon \to 0} \left(- 4 + 4 + 4 \epsilon + {\epsilon}^{2}\right)$
$= {\lim}_{\epsilon \to 0} \left(4 \epsilon + {\epsilon}^{2}\right) = 0$

We prove that ${\lim}_{x \to {2}_{-}} f \left(x\right) = f \left(- 2\right)$ and hence function is continuous at the point $x = - 2$.
$f ' \left(x\right)$ changes sign in $x = - 2$ and function has minimum value ${f}_{\min} = f \left(- 2\right) = 0$.

$f \left(2\right) = 0$
${\lim}_{x \to {2}_{+}} f \left(x\right) = {\lim}_{x \to {2}_{+}} \left(- 4 + {x}^{2}\right) = B$
$x = 2 + \epsilon$
$B = {\lim}_{\epsilon \to 0} \left(- 4 + {\left(2 + \epsilon\right)}^{2}\right) = {\lim}_{\epsilon \to 0} \left(- 4 + 4 + 4 \epsilon + {\epsilon}^{2}\right)$
$= {\lim}_{\epsilon \to 0} \left(4 \epsilon + {\epsilon}^{2}\right) = 0$

Again, we prove that ${\lim}_{x \to {2}_{+}} f \left(x\right) = f \left(2\right)$ and hence function is continuous at the point $x = 2$.
$f ' \left(x\right)$ changes sign in $x = 2$ and function has minimum value ${f}_{\min} = f \left(2\right) = 0$.