# 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:

- 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 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).**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).**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# **Domain:**

Clearly there is not any point for which#f(x)# is not defined.

Hence,#D_f=(-oo,oo)# **Range:**

#f(x)# cannot exceed the#y=2# . Also,#f(x)# cannot achieve#y=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]}