In the questions 7-11 we will linearize f(x) = va + 5 near a = 20. Question 7 1 pts To find a linearization of a function f() near point a, we need three things: a, f(a) and a slope f' (a). Then the linearization (tangent line) is L(x) = f' (a) (x - a) + f(a). Note this is a straight line with slope f' (a) passing through (a, f(a)). First we need to find f' (x): of' ( 20 ) = 2 3(x + 5) 3/2 of' (a) = 2Va +5 of' ( 2 ) = 1 2vac + 5 Of' ( 2 ) = - 1 Vac + 5 of' (2) = Va+5 Question 8 1 pts Now, a = 20 (given), of (a) = f (20 ) = 20 + 5 =5, f'(a) =f'(20) = 1 = V20 + 5 of(a) = f(20) = V20 + 5 =5, f'(a) =f'(20) =2120 +5 =10 of (a ) = f ( 20) = 20 + 5 = 5, f'(a) = f' (20) = 1 1 2120 + 5 10 Of(a) = f(20) = V20 + 5 = -5, f'(a) = f' (20) = 20 +5=5 Of(a) = f(20) = V20 + 5 =5, f'(a) = f'(20) = V20 +5 =5Question 9 1 pts This means that for a: near a. = 20, O x5z5(m20)+5 O w5z10(m5)+20 O :c20m5(a:20)5 1 ng{m5)+5 O O 1 \f$775m(a:720)+5 Question 10 1 pts Assume we need to estimate V 24.8. Note we only have an estimate for Val: + 5 for a: close to 20. Thus we write V 24. : V 19.8 + 5 and use the above estimate plugging in a: = 19.8, not 24.8. 0 V24. : V193 + 5 #10(19.8 5) + 20 O \f24. : 19.8 + x 509.8 7 20) + 5 o 1 1/24. : '/19.3 + z E(19.8 , 20) + 5 O 1/24. = M195 + m , (19.8 , 20) , 5 o 1 1/24. : $9.8 + m E095 , 5) + 5 D Question 11 1 pts We should all understand that the symbol ~ is very vague. Is it true that 7 ~ 3? Is it true that 109 ~ 115? The only way to make a statement containing ~ sensible is to give an error estimate. There is some actual value of the number v 19.8 and there is an estimate you found in the previous question. It can be proven that the error, the difference between actual value and the estimate, is bounded by |Error| S 2 2 ( 20 - a ) 2 . To bound the error, we need to find the second derivative of f (x) = vac + 5: it's f" (ac) = - (2 + 5)-3/2 Note a = 20 and x = 19.8. The error bound is O (20 - 5)-3/2 Error