Find the linearization at x = a ; y =(4+x) ^-1/2, a = 12

Answer 1

Let f(x)=(4+x)^-1/2. Now, the slope of the tangent line at any value of a, is f'(a). So, we find the derivative first.

Since f '(x)=-(1/2)(4+x)^-3/2, we get f '(12)=-(1/2)(16)^-3/2=-(1/2)(1/64)=-1/128. We also need the value of the y-coordinate when x=12. We plug into the original function, y=f(12)=16^-1/2=1/4. Now, we can use the point-slope formula to get

y-1/4=(-1/128)(x-12),

which simplifies to

y=(-1/128)x+9/32.