How do you graph, find the intercepts and state the domain and range of #f(x)=2-2^x#?

1 Answer
Jun 8, 2018

Answer:

Refer points below.

Explanation:

  1. We have to find out the Zeroes of the #f(x)# .
    #f(x)=2-2^x=0#
    #=>##2^x=2#
    #=>##x=1#
    We have one point on the graph is #(1,0)#
  2. First Derivative Test:
    We have find first derivative of #f(x)#.
    #f(x)=2-2^x#
    #f'(x)=-2^xln2#
    Note that #f'(x)# is always negative as#2^x>0# and #ln2>0#.
    #f'(x)<0# for #AA x inRR#
    #=># #f(x)#is always deceasing.-------(By First Derivative Test).
  3. Second Derivative Test:
    We have,
    #f(x)=2-2^x#
    #f'(x)=-2^xln2#
    #f''(x)=-2^x(ln2)^2#
    Note that #f''(x)# is always negative.
    #f''(x)<0# for #AA x inRR#
    #=># #f(x)#is always concave downward.-------(By second Derivative Test).
  4. Points on co-ordinate axes:
    We have already a point on #x#-axis i.e.#(1,0)# (as stated in point No. 1).
    We have to find out a point on #y#-axis.
    So, put #x=0# in #f(x)#
    #f(x)=2-2^x#
    #:.##f(0)=2-2^0#
    #:.##f(0)=2-1#
    #:.##f(0)=1#
  5. Domain:
    Clearly there is not any point for which #f(x)# is not defined.
    Hence, #D_f=(-oo,oo)#
  6. Range:
    #f(x)# cannot exceed the #y=2#. Also, #f(x)# cannot achieve #y=2#
    Hence, #R_f=(-oo,2)#
  7. Intercepts:
    #x#-intercept= 1 unit
    #y#-intercept= 1 unit
    We have found it by point No.4.
  8. Graph:
    By above Points we are able to draw the graph.
    graph{2-2^x [-10, 10, -5, 5]}