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

1 Answer
Jun 8, 2018

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 as2^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]}