The local approximation to the function g at x=1/2 is y=4x+1. What is the value of g(1/2) + g'(1/2)?

Given :

The local approximation to the function g at x = 1/2 is y = 4x+1.

To Find :

The value of g(1/2) + g'(1/2).

Solution :

It is given that at  g( y ) = 3x + 1 , at  x = 1/2.

Differentiating g( y ) w.r.t x ,we get :

g( y )' =  3

So, g( y ) + g( y' )  = 3/2 + 1 + 3

g( y ) + g( y' ) = 11/2

Hence, this is the required solution.

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph