How do you evaluate the function with the given values of #x: color(white)("d")f(x)=(1/2)[x+1] ;color(white)("ddd") x=0; x=-1#?

1 Answer
Tony B
Jun 22, 2018

#f(0)->y=1/2#
#f(-1)->y=0#

Explanation:

Suppose #x=0# then wherever there is an #x# write 0

#f(x)=1/2(x+1) color(white)("ddd")-> color(white)("ddd")f(0)=1/2(0+1)=1/2#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Suppose #x=-1# then wherever there is an #x# write -1

#f(x)=1/2(x+1) color(white)("ddd")-> color(white)("ddd")f(0)=1/2[color(white)(2/2)(-1)+1color(white)("d")]=0#

Related questions
See all questions in Function Notation
Impact of this question
2902 views around the world
You can reuse this answer
Creative Commons License