Question

1. f (s) = 23 – 4.x2 + 162 – 10 (a)Find the x- and y-coordinates of the critical point(s). Make a box around your answer. (b) Determine coordinates of the local maximum and minimum min if exist. If there is no local max/min, state this. Specify what Test you are using. Make boxes around your answers. Show all work. (c) Determine intervals of concavities up and down and the coordinates of the inflection point(s). If there is no inflection point, state this. Make boxes around your answers. (d) Use results in (a) and (b) to graph y(x). Plot the coordinates of the points found in (a)-(c) and the y-intercept. No credit if (a) - (c) not completed.

Answer 1

13 f(x)=_73 - 4x2 +168-10 for critical point f'(x) = 0 ) o ( 23 - 4x2 +16X-10) = 0 da 3 = 3x² - ax + 16 =0 3 x² - 6x + 16 = 0 81 6012- 4x4X16. a = a + / 64-64. 2 n=4 This is f14) E only one critical point _- 4.42 +16x4'-10 34 +14) = 54-60 Hry > +34

I Date: أمهل for extremum ( Double derivative test) = f"(x) Tit f(x) > 2x1-8_ Minima l Lit f"(x) < (Maxima At critical point x = 4 Little {"(4) = 2x4-8 inflection f" (4) -01 There is no local Minima & Maxima But there is inflection point I point where Concavity changes) - fil(a) = 23-d iff"!24) >0 { Concare upwards - 2a-do 2x> [ x4] Concave downward) f"(a) co 2x-2 2 < <o 4

Concave down Concave upwards f'(x) > 0 f"(m)< o - At x=4 (inflection point) f(t) = 34 - 11.333 my 11.332 point of inflection (0.76) 4- 1(0-10)