Is f(x)=-7x^3+x^2-2x-1 increasing or decreasing at x=-2?

Jun 3, 2018

See explanation.

Explanation:

To find if a function is increasing or decreasing at a given point we have to calculate its derivative at the point.

The function is:

$f \left(x\right) = - 7 {x}^{3} + {x}^{2} - 2 x - 1$

so its derivative is:

${f}^{'} \left(x\right) = - 21 {x}^{2} + 2 x - 2$

The value of derivative at ${x}_{0} = - 2$ is:

${f}^{'} \left(2\right) = - 21 \cdot {\left(- 2\right)}^{2} + 2 \cdot \left(- 2\right) - 2 = - 84 - 4 - 2 = - 90$

The value of the derivative is less than zero, so the function is decreasing.

Jun 3, 2018

Either drawing the graph or taking the derivate $f ' \left(x\right) = - 21 {x}^{2} + 2 x - 2$ in $x = - 2$ we find that the slope is negative ($f ' \left(- 2\right) = - 90$). We can, therefore, conclude that $f \left(x\right)$ is decreasing at $x = - 2$

Explanation:

Let's start by drawing the graph in Geogebra (an excellent, free geometry program): Looking at it , it is quite clear that $f \left(x\right)$ is decreasing at $x = - 2$

We can also find this without drawing the graph by taking the derivate, which gives the slope of the tangent in each value of x:
$f ' \left(x\right) = - 7 \cdot 3 {x}^{2} + 2 x - 2 = - 21 {x}^{2} + 2 x - 2$
Insert the value $x = - 2$:
$f ' \left(- 2\right) = - 21 {\left(- 2\right)}^{2} + 2 \left(- 2\right) - 2 = - 84 - 4 - 2 = - 90$

As $f ' \left(- 2\right) = - 90 < 0$, we can conclude that $f \left(- 2\right)$ is decreasing in -2.