Question

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

Answer 1

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]}