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:
- 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) - First Derivative Test:
We have find first derivative off(x) .
f(x)=2-2^x
f'(x)=-2^xln2
Note thatf'(x) is always negative as2^x>0 andln2>0 .
f'(x)<0 forAA x inRR
=> f(x) is always deceasing.-------(By First Derivative Test). - Second Derivative Test:
We have,
f(x)=2-2^x
f'(x)=-2^xln2
f''(x)=-2^x(ln2)^2
Note thatf''(x) is always negative.
f''(x)<0 forAA x inRR
=> f(x) is always concave downward.-------(By second Derivative Test). - Points on co-ordinate axes:
We have already a point onx -axis i.e.(1,0) (as stated in point No. 1).
We have to find out a point ony -axis.
So, putx=0 inf(x)
f(x)=2-2^x
:. f(0)=2-2^0
:. f(0)=2-1
:. f(0)=1 - Domain:
Clearly there is not any point for whichf(x) is not defined.
Hence,D_f=(-oo,oo) - Range:
f(x) cannot exceed they=2 . Also,f(x) cannot achievey=2
Hence,R_f=(-oo,2) - Intercepts:
x -intercept= 1 unit
y -intercept= 1 unit
We have found it by point No.4. - Graph:
By above Points we are able to draw the graph.
graph{2-2^x [-10, 10, -5, 5]}