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

Question

Isaiah Anthony · Accepted Answer

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

Genesis Allen · Answer

Range:
#1 <= y < oo#

#y in [1,oo)# 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.

Isaiah Anthony · Answer

undefined The range of the function $ f(x) = x^2 + 2x + 2 $ is all real numbers greater than or equal to the minimum value of the function. Since the function represents a quadratic with a positive leading coefficient, it opens upwards, and its minimum value occurs at the vertex. The range is therefore all real numbers greater than or equal to the y-coordinate of the vertex.