Question

6. For a certain function f(x) we have: f'(x) = (x - 3)²(2x - 3) and • f"(x) = 6(x - 3)(x - 2) (a) Use f' to find the intervals where f is increasing, the intervals where f is decreasing, the x- coordinates and nature (max, min or neither) of any local extreme values. (b) Use f" to find the intervals where the graph of f is concave up, the intervals where the graph of f is concave down and the x-coordinates of any inflection points on the graph of f. (c) Suppose the point (1,3) is on the graph of f. Find the slope-intercept (y = mx + b) equation of the tangent line to the graph of f at this point.

Answer 1

Quey 4/2) = (2+ 37² (2x-3) {"() = 662-3)(2-2) Anicos Analysic f'(x) = (x-33262x<3) 42-33320 I, (2x-3) 20 p=2 x 3 & x= 3 are critical Prints _30239 2 2 + 3 2 1 312 Take any value f(x) Sign function Interucel Behaviour (- 00 3/2) 0 flaixo Decreasing | ( 312,3) f(x) > Increating (3.960) 4 f(x)ro Forrealing Interval of Increase (312, 3) U (30) Interval af Decreare (-60, 342) wohen fundion moves from increase to decrease about a poine then that point is Local Maximum while decrease to increase then Local Minimum there Local Toaxim um is at x= No Poins Local Minimum is at a=312 (6) Now f"20) 6(x3)(x+2) = 0 X-320 x-2=0 13223 122 So, Point of Inflections are a=3 & x=2 take any value het any I" (x) Sign Interval C-c092) (23) (3,0) O 2.5 4 function Behaviwy f"(x) 70 Concave up t"(x) to Concane Down flairo Concave up

(قل) لم مملحه Internal of Concave up = (-09,2) U13,0) Interval of Concave Down = (2,3) (4) kausing pan (4.4) - 143) f (x) = (2-3) ² (2x-3) J1W=(1-3)2 (26-3) = 24 Slope. mar4 Equation of Tanglent me with fouing pointe Gay) = (43) and slope m=-4 ly-y) = m ( x-24) => (y - 3) = (-6(x-1) -42th yo -4847

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