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

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

Find the linearization L(x) of the function at a.f(x) = cos(x), a = 3𝜋/2

Find the linearization L(x) of the function at a.f(x) = sin(x),    a = 𝜋3

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 of the function f(x) = x + 1 at a = 8 and use it to approximate the numbers 8.95 and 9.04. Are these approximations overestimates or underestimates?SolutionThe derivative of f(x) = (x + 1)1⁄2 isf ′(x) = 12​(x+1)−(12​) ,and so we have f(8) = and f ′(8) = . Putting these values into the equation L(x) = f(a) + f ′(a)(x − a), we see that the linearization isL(x)= f(8) + f '(8)(x − 8) = +  (x − 8)= .The corresponding linear approximation isx + 1 ≈ + x6        (when x is near 8).In particular, we have8.95 ≈ 53 + 6 =     (round to four decimal places)and  9.04 ≈ 53 + 6 =     (round to four decimal places).