Answer:
Lf(x) = Lg(x) = Lh(x) = 1 - 2x
value of the functions and their derivative are the same at x = 0
Step-by-step explanation:
Given :
f(x) = (x − 1)^2,
g(x) = e^−2x ,
h(x) = 1 + ln(1 − 2x).
a) Determine Linearization of f, g and h at a = 0
L(x) = f (a) + f'(a) (x-a) ( linearization of <em>f</em> at <em>a</em> )
<u>for f(x) = (x − 1)^2 </u>
f'(x ) = 2( x - 1 )
at x = 0
f' = -2
hence the Linearization at a = 0
Lf (x) = f(0) + f'(0) ( x - 0 )
Lf (x) = 1 -2 ( x - 0 ) = 1 - 2x
<u>For g(x) = e^−2x </u>
g'(x) = -2e^-2x
at x = 0
g(0) = 1
g'(0) = -2e^0 = -2
hence linearization at a = 0
Lg(x) = g ( 0 ) + g' (0) (x - 0 )
Lg(x) = 1 - 2x
<u>For h(x) = 1 + ln(1 − 2x).</u>
h'(x) = -2 / ( 1 - 2x )
at x = 0
h(0) = 1
h'(0) = -2
hence linearization at a = 0
Lh(x) = h(0) + h'(0) (x-0)
= 1 - 2x
<em>Observation and reason</em>
The Linearization is the same in every function i.e. Lf(x) = Lg(x) = Lh(x) this is because the value of the functions and their derivative are the same at x = 0
