Let f(x) = (x − 1)2, g(x) = e−2x, and h(x) = 1 + ln(1 − 2x). (a) Find the linearizations of f, g, and h at a = 0. What do you notice? How do you explain what happened?

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 f at a )

for f(x) = (x − 1)^2  

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

For g(x) = e^−2x

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

For h(x) = 1 + ln(1 − 2x).

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

Observation and reason

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

which expression represents the approximate length of bc?

step-by-step explanation:

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5