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

answer: 370

step-by-step explanation:

remember use pemdas! there are no parentheses or exponents so lets move onto the first part of the equation: multiplication and division. first, lets break the first part of the equation up into something easier using the distributive property. 28 * 12= (20 *12) + (8*12) you know that 20*12 is 240 and 8*12 is 96. add them and you get 336. then, lets solve the last part of the equation (the last part of pemdas is addition and subtraction). 336 plus 34 is 370. hope this !

answer: im gonna go with milli

step-by-step explanation: