What is the domain of the function # f(x) = (x+1)(x^2-x+2)#?

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Answer 1 · 2018-01-02T01:42:11

For #f(x) = (x+1)(x^2-x+2)# See answer by @Steve M above.

However, I prepared an answer to

#f(x) = (x+1)/(x^2-x+2)# which I will retain for interest

#f(x)# is defined #forall x in RR: (x^2-x+2) != 0#

Consider: #x^2-x+2 = 0#

Here we have a quadratic, where the discriminant

#Delta = -1^2 -4*1*2 =-7#

Since #Delta < 0# the equation has no real roots.

Hence, #f(x)# is defined #forall x in RR#

#:.# the domain of #f(x)# is #(-oo, +oo)#

Answer 2 · 2018-01-02T01:43:25

#D_f=RR#

#f(x)=(x+1)/(x^2-x+2)#

In order for this function to be defined in #RR# we need,
#x^2-x+2!=0#

Let's solve the #=# to find the roots of the equation.

We have #x^2-x+2=0#

Discriminant of this quadratic equation is negative which means

#x^2-x+2# #>0# , #x##in##RR#

Therefore the domain of #f# is #RR#

Answer 3 · 2018-01-02T19:35:55

The domain of #f# is #RR#

We have:

# f(x) = (x+1)(x^2-x+2) #

This is the product of two polynomials, both of which themselves have a domain of #RR#

hence the domain of #f# is also #RR#