# What are the critical points of  f(x)=3x^4+6x^3?

Nov 14, 2015

$12 {x}^{3} + 18 {x}^{2}$

#### Explanation:

You need to compute the first derivative, and then to find its zeroes. We need three simple rules:

• The derivative of a power ${x}^{n}$ is $n \cdot {x}^{n - 1}$
• Scalar constats can be ignored if they multiply: $\left(k f \left(x\right)\right) ' = k \cdot f \left(x\right) '$
• The derivative of a sum is the sum of the derivatives.

The third rules means that we can compute the two derivatives separately and then sum them, while the first and the second mean that we have

$\frac{d}{\mathrm{dx}} 3 {x}^{4} = 3 \cdot 4 {x}^{3} = 12 {x}^{3}$, and

$\frac{d}{\mathrm{dx}} 6 {x}^{3} = 6 \cdot 3 {x}^{2} = 18 {x}^{2}$