Suppose the linear approximation for a function f(x) at a = 2 is given by the tangent line y = −2x + 10.What are f(2) and f '(2)?f(2) = f '(2) = If g(x) =[f(x)]2, find the linear approximation for g(x) at a = 2.L(x) =

Suppose that we don't have a formula for g(x) but we know that g(1) = −5 and g'(x) = x2 + 8 for all x.Use a linear approximation to estimate g(0.9) and g(1.1).g(0.9) ≈ g(1.1) ≈

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).

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.

Recall that the slope m of the graph of f at the point (x, f(x)) is equal to the slope of its tangent line at (x, f(x)), and is given by the following provided this limit exists.m = lim h→0 msec = lim h→0 f(x + h) − f(x)hTherefore, we will first need to find the expression that represents the secant line at the point (2, −10).First, set up the difference equation for the point (2, −10) by substituting x = 2.m  =  lim h→0 f(x + h) − f(x)h =  lim h→0 f  + h − f  h

Find the linear approximation of the function g(x) = 51 + x at a = 0.