so we have f(3) = and f '(3) = . Putting these values into this equation, we see that the linearization isL(x)= f(3) + f '(3)(x − 3) = + (x − 3)= .

Find the linearization of the function f(x) = x + 6 at a = 3 and use it to approximate the numbers 8.99 and 9.01. Are these approximations overestimates or underestimates?SOLUTION The derivative of f(x) = (x + 6)1/2 isf '(x) = (12​)(x+6)(−(12​))·1 .and so we have f(3) = and f '(3) = . Putting these values into this equation, we see that the linearization isL(x)= f(3) + f '(3)(x − 3) = + (x − 3)= (16​)x+172​ .The corresponding linear approximation isx + 6 ≈ + x6        (when x is near 3).In particular, we have8.99 ≈ 52 + 6 =     (round to four decimal places)and  9.01 ≈ 52 + 6 =     (round to four decimal places).The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 3. We also see that our approximations are overestimates because the tangent line lies above the curve.     Of course, a calculator could give us approximations for 8.99 and 9.01, but the linear approximation gives an approximation over an entire interval.

Find the linearization L(x) of the function at a.f(x) = x3/4, a = 16

Which of the given function is linear?*f (x) = | 3x – 1|f(x) = 2/(x + 7)f(x) = √x - 4f(x) = 2x/3

Find the linearization L(x) of the function at a.f(x) = x4 + 2x2, a = −1

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