Question

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

Answer 1

Perhaps this question should be in the calculus section.

Domain`-> color(white)(..)-oo < x < +oo`

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`->`d for domain
output`->`r for range

`color(brown)("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'.

`ul("As no such condition exists")` we may use any value we so wish for `x` such that `-oo < x < +oo`

`color(white)()`

`color(brown)("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 `3x^2` making it even greater in influence over `-4x^3`

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

`lim_(x->+-oo) y = lim_(x->+-oo) (x^4-4x^3+3x^2)->k=+oo`

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

To answer this I am choosing to use calculus.

Set `y=x^4-4x^3+3x^2`

Then `dy/dx=4x^3-12x^2+6x`

Set `" "4x^3-12x^2+6x=0`

`x(4x^2-12x+6)=0`

`x=0` is one.

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

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

I will let you finish this off.