# Is f(x)=xe^(x^3-x)-1/x^3 increasing or decreasing at x=4?

$f \left(x\right)$ is INCREASING at $x = 4$

#### Explanation:

the given equation is $f \left(x\right) = x \cdot {e}^{{x}^{3} - x} - \frac{1}{x} ^ 3$

Let us solve first the first derivative then substitute the value $x = 4$

If $f ' \left(4\right) =$positive value, then it is INCREASING at $x = 4$

If $f ' \left(4\right) =$negative value, then it is DECREASING at $x = 4$

If $f ' \left(4\right) =$zero value, then it is STATIONARY at $x = 4$

First derivative

$f ' \left(x\right) = x \cdot \frac{d}{\mathrm{dx}} \left({e}^{{x}^{3} - x}\right) + {e}^{{x}^{3} - x} \cdot \frac{d}{\mathrm{dx}} \left(x\right) - \frac{d}{\mathrm{dx}} \left({x}^{- 3}\right)$

$f ' \left(x\right) = x \cdot \left({e}^{{x}^{3} - x}\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{3} - x\right) + {e}^{{x}^{3} - x} \cdot 1 - \left(- 3\right) \left({x}^{- 3 - 1}\right)$

$f ' \left(x\right) = x \cdot \left({e}^{{x}^{3} - x}\right) \cdot \left(3 {x}^{2} - 1\right) + {e}^{{x}^{3} - x} + 3 \left({x}^{- 4}\right)$

Evaluate at $x = 4$

$f ' \left(4\right) = 4 \cdot \left({e}^{{4}^{3} - 4}\right) \cdot \left(3 {\left(4\right)}^{2} - 1\right) + {e}^{{4}^{3} - 4} + 3 \left({4}^{- 4}\right)$

$f ' \left(4\right) = 4 \cdot \left({e}^{{4}^{3} - 4}\right) \cdot \left(3 {\left(4\right)}^{2} - 1\right) + {e}^{{4}^{3} - 4} + 3 \left({4}^{- 4}\right)$

$f ' \left(4\right) = \text{positive value}$

Therefore INCREASING at $x = 4$

God bless....I hope the explanation is useful.