Performing the second approximation, ,x3 = x2 – [( x1 – x2) / (f(x1) – f(x2))]f(x2)=(- 0.234375) – [(1 – 0.25)/(-3 – (- 0.234375))](- 0.234375)= 0.186441Hence, f(x3) = 0.074276

Rewrite the function f(x)=0.25(0.5)–3x in the form f(x)=a(b)x.

x1=0.35*(x1)+0.37*(x2)+0.19*(x3) x2=0.01*(x1)+0.11*(x2)+0.15*(x3)+0.05*(x4) x3=0.01*(x2)+0.1*(x3)+0.02*(x4) x4=0.64*(x1)+0.51*(x2)+0.56*(x3)+0.93*(x4) x1+x2+x3+x4=1 solving equations

Find the linear and quadratic approximations to the following function about x = 0 :f (x) = (1 + x)5and so find approximations for 1.055

Evaluate the function f(x) at the given numbers. (Round your answers to six decimal places.)f(x) = 5ex − 5 − 5xx2, x = 1, 0.5, 0.1, 0.05, 0.01, −1, −0.5, −0.1, −0.05, −0.01f(1) = f(0.5) = f(0.1) = f(0.05) = f(0.01) = f(−1) = f(−0.5) = f(−0.1) = f(−0.05) = f(−0.01) = Guess the value of the limit of f(x) as x approaches 0. (Round your answer to six decimal places. If an answer does not exist, enter DNE.)lim x→0 5ex − 5 − 5xx2 =

Step 4: Repeat the process using x1 and x2: f(x2) = f(-2) = (-2)^3 - 5*(-2) + 1 = -8 + 10 + 1 = 3 x3 = x2 - f(x2)((x2 - x1) / (f(x2) - f(x1))) x3 = -2 - 3((-2 - 1) / (3 - (-3))) x3 = -2 - 3*(-3) x3 = -2 + 9 = 7