Question

9. Find the equation for the line that is tangent to the graph of f (z) = (92 - 1) at the value 3 1.

Answer 1

Soli Given, f(x) = (92-1013 we need to find the equation of the tansent line at x=1 To find the equation of line, we should have two points on the line Cor) one point and slope of the line. Here, in this problem, we can have one point and slope to find the equation of the tensent line - point is? at x=1 & f(1) = (acu-u" (9-1543 8-13 8/3² (23) 는 th e point is: (1/₂) a slope is: df(x)/2=1 da ( 92–1713] = (12-1750) 3 [: Chain rule: df (90x)) dx fo)).g.) t-3 (97-15 4/2 at x=1 df(x) da 5) dou) do 2-1 1/2-1 -413 =-3 (961)-1) = -3 (8) 1413 = -3 (34) --> (225) slope = -3/16

Sol equation of the tensent line is: By slope - point form of the line slope m and poina (01 Jo ine is : (Y-yo) = m(-1) here, m=-3/16 1(oiyo) = L 2) equation is: (4-4/2) = 3 16 (-1) = 키 y일 디저름을 휴 나 16 3 16 이y 3 16. 3 것 커 8 i6 16 15 (