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How do you find the domain and range of #f(x) = x^2 + 5#?

Answer 1

Savannah Anderson

Domain: #(-oo,+oo)#
Range: #[5,+oo)#

#f(x) = x^2+5#

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

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

#f(x)# is a parabola and hence has a single critical value.

Since the coefficient of #x^2# is positive, #f(x)# has an absolute minimum value and no upper bound.

Since #x^2 >=0 forall x in RR -> f(x)_"min" = f(0)#

#:. f(x)_"min" = 5#

Hence, the range of #f(x)# is #[5,+oo)#

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Answer 2

Avery Alexander

To find the domain of ( f(x) = x^2 + 5 ), note that it is a polynomial function, which is defined for all real numbers. So, the domain is all real numbers, or (\mathbb{R}).

To find the range, notice that the minimum value of ( x^2 ) is 0, since any real number squared is non-negative. Adding 5 to this minimum value gives 5. So, the range is all real numbers greater than or equal to 5, or ([5, \infty)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.