# What are the absolute extrema of f(x)=5x^7 - 7x^5 - 5 in[-oo,oo]?

Jan 26, 2016

There are not absolute extrema because $f \left(x\right)$ unbounded
There are local extrema:

LOCAL MAX: $x = - 1$
LOCAL MIN: $x = 1$
INFLECTION POINT $x = 0$

#### Explanation:

There are not absolute extrema because

${\lim}_{x \rightarrow \pm \infty} f \left(x\right) \rightarrow \pm \infty$

You could find local extrema, if any.

To find $f \left(x\right)$ extrema or critical poits we have to computate $f ' \left(x\right)$

When $f ' \left(x\right) = 0 \implies f \left(x\right)$ has a stationary point (MAX, min or inflection point).
Then we have to find when:

$f ' \left(x\right) > 0 \implies f \left(x\right)$ is increasing
$f ' \left(x\right) < 0 \implies f \left(x\right)$ is decreasing

Therefore:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(5 {x}^{7} - 7 {x}^{5} - 5\right) = 35 {x}^{6} - 35 {x}^{4} + 0 = 35 {x}^{4} \left({x}^{2} - 1\right)$

$\therefore f ' \left(x\right) = 35 {x}^{4} \left(x + 1\right) \left(x - 1\right)$

• $f ' \left(x\right) = 0$

$\textcolor{g r e e n}{\cancel{35}} {x}^{4} \left(x + 1\right) \left(x - 1\right) = 0$

${x}_{1} = 0$
${x}_{2 , 3} = \pm 1$

• $f ' \left(x\right) > 0$

${x}^{4} > 0$ $\forall x$
$x + 1 > 0 \implies x > - 1$
$x - 1 > 0 \implies x > 1$

Drawing the plot, you'll find

$f ' \left(x\right) > 0 \forall x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$

$f ' \left(x\right) < 0 \forall x \in \left(- 1 , 1\right)$

$\therefore f \left(x\right)$ increasing $\forall x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$
$\therefore f \left(x\right)$ decreasing $\forall x \in \left(- 1 , 1\right)$

$x = - 1 \implies$LOCAL MAX
$x = + 1 \implies$ LOCAL MIN
$x = 0 \implies$ INFLECTION POINT

graph{5x^7-7x^5-5 [-16.48, 19.57, -14.02, 4]}

Jan 26, 2016

${\lim}_{x \rightarrow \infty} f \left(x\right) = \infty$ and ${\lim}_{x \rightarrow - \infty} f \left(x\right) = - \infty$.