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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#?

Answer 1

Julia Adams

Evaluate ( f(x) = \frac{1}{2}(x + 1) ) for ( x = 0 ) and ( x = -1 ):

Substitute ( x = 0 ): ( f(0) = \frac{1}{2}(0 + 1) = \frac{1}{2} )
Substitute ( x = -1 ): ( f(-1) = \frac{1}{2}(-1 + 1) = 0 )

Therefore, ( f(x) = \frac{1}{2}(x + 1) ) evaluates to ( \frac{1}{2} ) when ( x = 0 ) and to ( 0 ) when ( x = -1 ).

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Answer 2

Eva Adamson

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

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#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.