Question

is: 6. (8 points) / is a function that is continuous on (-0,00). The first derivative of /"(x) = (3x - 1)x+3X5 - x) Use this information to answer the following questions about : a. On what intervals is increasing or decreasing? Internal in which fis increasing or -- 8x-1) (x+3)(5-x) > 0 x=112, -3, -5 b. At what values of x does f have any local maximum or minimum values? - V2 ; Location(s) of Minima: Location(s) of Maxima: 7. (10 points) f is a function that is continuous on (-0,00). The first derivative of f is: S'(x)=(x-3)(x + 2). Use this information to find: a. The intervals on which is concave upward or downward. finflection of the granh nf Answer:

Answer 1

By HomeworkLib guidelines, 1st question is enough to answer. So that I am answering first one, plz post the other questions separately. 6. Given f'(x) = (3x – 1)(x+3)(5 – x) (a) Solve f'(x) = 0) = (3x – 1)(x+3)(5 – x) = 0 >> 3x – 1=0 or x +3 = 0 or 5- x = 0 => critical points off : x=ş,x= -3,x = 5 = domain off is divided into subintervals: (-2,-3) (-3,4) (5)(5,

= ifxe(-2,-3)+(-3) (4:5 )u(3,9) then xe(-,-3)or xel - 35) or xe (5)or xe(5.5 We know that f is increasing if and only if f'(x) > 0 = (3x – 1)(x+3)(5 – x) > 0 If x = (-00,-3), take test point x = -4 € (-0, -3), then f'(-4)= (3(-4) – 1)(-4+3)(5-(-4))=117>0 >f is increasing on (-0, -3)

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Ifxe(-3.), take test point x = 0 € (-3.), then f'(O) = (3(0) – 1)(0+3)(5 – 0) = -15<0 = fis decreasing on (-35) Ifxe( 5 ) take test point x = 1e( 5 ) then f'(1)= (3(1) – 1)(1+3)(5 – 1) = 32 > 0 =f is increasing on (15)

If x € (5,0), take test point x = 6 € (5,00), then f'(O) = (3(6) – 1)(6+3)(5-6)=-153< 0 = f is decreasing on (5,00) :: f is increasing on the intervals (- f is decreasing on the intervals

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(6) Since f changes as increase to decrease at x = -3, x = 5, f local maximum at x = -3, x = 5 Since f changes as indecrease to increase at x = = f local minimum at x = 1

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