What is the range of the function #f(x)=x^2+2x+2#?

2 Answers
Apr 6, 2017

Answer:

#{y|yge1}# or #y in [1, oo[#

Explanation:

The function can be rewritten as #f(x)=x^2+2x+1+1=(x+1)^2+1#. Notice that #(x+1)^2≥0# for all #x in RR#. Thus, the values #f(x)# can be is all values greater than or equal to #1#.

Thus, the range is #{y|yge1}# or #y in [1, oo[#.

Apr 6, 2017

Answer:

Range:
#1 <= y < oo#

#y in [1,oo)#

Explanation:

Since this is an upward facing parabola, we know that the upper bound of the range is #+oo#

Find the minimum. We can do this by either using our knowledge of parabolas and quadratic form to complete the square to find the vertex, and the y value is the minimum; OR we could use calculus and find where #f'(x)# is equal to zero, and find the corresponding #y# value.

#color(blue)("Method 1: Vertex form"#
#f(x)=x^2+2x+2#

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

Vertex is at #(-1,1)#, so lowest #y# value is #1#.

#color(red)("Method 2: Set derivative equal to zero"#
#f(x)=x^2+2x+2#

#f'(x)=2x+2#

#0=2x+2#

#x=-1#

#f(-1)=(-1)^2+2(-1)+2=1#
The minimum is at #x=-1#, and #f(-1)=1# so #y= 1# is the lowest y value.