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

Question

Eva Adamson · Accepted Answer

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.

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.

Ethan Adkins · Answer

undefined To find the domain of the function $f(x) = x^4 - 4x^3 + 3x^2$, we note that it is a polynomial function, which is defined for all real numbers. Therefore, the domain is all real numbers.

To find the range of the function, we can analyze the behavior of the function. Since $x^4$, $-4x^3$, and $3x^2$ are all non-negative for all real numbers $x$, the function $f(x)$ will always be non-negative or zero. Therefore, the range of $f(x)$ is $[0, +\infty)$.