# Question 072cb

May 17, 2017

To find the slope, evaluate $f ' \left(0\right)$:

$f ' \left(0\right) = - 2$

#### Explanation:

The equation for the slope of the tangent line is:

$f ' \left(x\right) = 5 {x}^{4} + 3 {x}^{2} - 2$

We can find local maxima by computing the next derivative, setting that equal to 0, and then solving for the value(s) of x.

Compute the next derivative:

$f ' ' \left(x\right) = 20 {x}^{3} + 6 x$

Set it equal to 0:

20x^3+ 6x = 0

Factor:

$2 x \left(10 {x}^{2} + 3\right) = 0$

$x = 0 \mathmr{and} x = \pm \sqrt{\frac{3}{10}} i$

$x = 0$

Compute the next derivative:

$f ' ' ' \left(x\right) = 60 {x}^{2} + 6$

Evaluate it at $x = 0$:

$f ' ' ' \left(0\right) = 60 {\left(0\right)}^{2} + 6 = 6$

$6 > 0$, therefore, the slope is a local minimum at $x = 0$

To find the slope, evaluate $f ' \left(0\right)$:

$f ' \left(0\right) = - 2$

May 17, 2017

$- 2$

#### Explanation:

The tangent line will be the derivative of the function.
$f ' \left(x\right) = 5 {x}^{4} + 3 {x}^{2} - 2$

To find the minimum value of this line we have to find the critical points. So we find the second derivative
$f ' ' \left(x\right) = 20 {x}^{3} + 6 x$

The minimum will be when this function is equal to zero
$0 = 20 {x}^{3} + 6 x$
$0 = x \left(20 {x}^{2} + 6\right)$

From this we can see that either factor will only be equal to zero when $x = 0$.

So we plug this into the tangent line formula to get the slope at this point.
$f ' \left(0\right) = 5 {\left(0\right)}^{4} + 3 {\left(0\right)}^{2} - 2$
$f ' \left(0\right) = 0 + 0 - 2$
$f ' \left(0\right) = - 2$