# How do you find the domain and range of f(x)=(x^4)-(4x^3)+3x^2?

Jul 18, 2017

Perhaps this question should be in the calculus section.

Domain$\to \textcolor{w h i t e}{. .} - \infty < x < + \infty$

I have taken you to a point where you can finish off for the range.

#### Explanation:

Input comes before you can get any output.

As a memory aid: d for domain comes before r for range so the link is:

input$\to$d for domain
output$\to$r for range

$\textcolor{b r o w n}{\text{Determine the domain}}$

There are no denominators so no 'excluded' values

If you have a variable in the denominator and it has the 'ability' to 'turn' the denominator into 0 then we have a problem.

YOU ARE NOT ALLOWED TO DIVIDE BY 0

Thus the expression/equation becomes 'undefined'.

$\underline{\text{As no such condition exists}}$ we may use any value we so wish for $x$ such that $- \infty < x < + \infty$

$\textcolor{w h i t e}{}$

$\textcolor{b r o w n}{\text{Determine the range}}$

As the value of $x$ becomes more and more positive then the influence of ${x}^{4}$ becomes more and more influential. Not only that, it is added to by the $3 {x}^{2}$ making it even greater in influence over $- 4 {x}^{3}$

Note that if $x < 0$ then ${x}^{4} > 0 \mathmr{and} {x}^{2} > 0$. Not only that but $- 4 {x}^{3}$ is also positive. Consequently for $x < 0$, $y$ grows even faster.

${\lim}_{x \to \pm \infty} y = {\lim}_{x \to \pm \infty} \left({x}^{4} - 4 {x}^{3} + 3 {x}^{2}\right) \to k = + \infty$

Ok that has dealt with the maximums but what about the minimums.

To answer this I am choosing to use calculus.

Set $y = {x}^{4} - 4 {x}^{3} + 3 {x}^{2}$

Then $\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} - 12 {x}^{2} + 6 x$

Set $\text{ } 4 {x}^{3} - 12 {x}^{2} + 6 x = 0$

$x \left(4 {x}^{2} - 12 x + 6\right) = 0$

$x = 0$ is one.

Solve $4 {x}^{2} - 12 x + 6 = 0$ as a normal quadratic.

The values of $y$ can be found by substitution.

I will let you finish this off.