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

1 Answer
Jul 18, 2017

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.

Explanation:

Tony B

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.