1−x−−−−√≈−12x+1
0.9−−−√=1−0.1−−−−−−√≈−12(0.1)+1
To simplify:
−1/2⋅1/100+1=−1/2⋅1/100+200/200=200−1/200=199/200 or =.995
So yes for sqrt (.99), that is a good approximation.
To write linear approximations:
Our curve should be y=f(x) and substitute x=a for the numbers m
Our formula for the
linear equation is y=mx+b
So subtsitute
y′=f′(x)
y=f′(a)x+b
Then find the y-intercept, which is b.
Given: (a, f(a))=
point of the line
Then substitute, that would be
f(a)=f′(a)⋅a+b
f(a)−f′(a)⋅a=b
So the tangent line to y=f(x) at x=a
Or the equation we will use for linear approximations is:
y=f′(a)x+f(a)−f′(a)⋅a
y=f′(a)⋅(x−a)+f(a)
