Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The linear approximation to ?f?(?x?)= ?x?2 at the point (0, 0) is ?L?(?x?) = 0 b. Linear approximation provides a good approximation to ?f?(?x?)= | ?x? |at(0, 0). c. If ?f?(?x?) = ?mx? + ?b,? then at any point ?x? = ?a.? the linear approximation to ?f? is ?L?(?x?)= f?(?x?).

Solution 39E STEP 1 2 (a). The linear approximation to f(x) = x at the point (0, 0) is ) = 0. Let f be a function differentiable at an interval containing a point a.The linear approximation to f at a is given by L(x) = f(a)+f (a)(xa) Given f(x) = x and (a,f(a)) = (0,0) f(a) = 0. f(x) = 2x f (a) = 0 Therefore L(x) = 0 Thus the given statement is true. STEP 2 (b). Linear approximation provide s a d approximation to f (x)= | x | at(0, 0) .The linear approximation to f at a is given by L(x) = f(a)+f (a)(xa) Here f(x) = |x| is not differentiable at x=0. Therefore f(a) does not exist. Therefore the given statement is false. STEP 3 (c). If f(x) = mx + b, then at any point x = a. the linear approximation to f is L(x)= f(x). Given f(x) = mx+b and x=a. .The linear approximation to f at a is given by L(x) = f(a)+f (a)(xa) f(a) = ma+b Therefore f(x) = m f(a) = m Therefore L(x) = ma+b+m(xa) = ma+b+mxma = mx+b Thus the given statement is true,ie, L(x) = f(x)