The local linear approximation to the function #f# at #x=1# is #y=2x+8# What is the value of #f(1)+f'(1)#?

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Answer 1 · 2017-10-30T16:46:49

#f(1)+f'(1) = 12#

If #y=f(x)# is a differentiable function, then the linear approximation around #x=1# is:

#f(x) ~= f(1) +f'(1)(x-1) = f'(1)x + f(1)-f'(1)#

Comparing with the linear approximation #y=2x+8# we get:

#f'(1)x + f(1)-f'(1) = 2x+8#

and equating the coefficient of the same degree in #x#:

#f'(1) = 2#

#f(1)-f'(1) = 8 => f(1) = 10#

So:

#f(1)+f'(1) = 12#

The local linear approximation to the function #f# at #x=1# is #y=2x+8# What is the value of #f(1)+f'(1)#? ​