Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3| <= ?

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: