I don\'t know if these answers are accurate. CALCULUS I-1950 1. The graph below

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I don't know if these answers are accurate.

CALCULUS I-1950 1. The graph below is the graph of f', the derivative of f The domain of the derivative is -5Srs6 5 rtx) The critical points forfarex=43-1+1+2+3-4 The critical points for f" are32 12 3 1) 2) 15, *3) fhas a local maximum when x_3 *4) 5) 6) 7) 8) f has its maximum value on [-5,6] when x fis decreasing on the interval(s) The graph of f is concave up on the interval(s)_ $53 The x-coordinates of the points of inflection are 5,1) (3.61 (12) -3-1 x = t 2. ar, has its maximum value when x = *9) f has its maximum value when Does f, have a minimum value on [-5,6)? Explain ves,f(x) chers-tr bedre Does " have a minimum value on [-5,61? Explain yes ol Ceaust ran nak re t 10) * 11) otrve Prom

Explanation / Answer

1. Critical points for f = -3, 1, 3 ( when f'(x) is zero)

2. Critical points for f' = -3,-1,2, 4, 5 ( when f'(x) has local minima, local Maxima and inflection point.)

3. f has local maximum = 3 ( when f' going from positive to negative.)

4. f has maximum value at x= -5 ( because f' has more negative part than positive part before tha end of positive interval of f'(x) )

5. Decreasing interval = [-5,1] and (3, 6] ( for decreasing f' should be negative)

6. f concave up = [ -5, -3) and (-1, 2) ( when f' is increasing)

7. Inflection point of f = -3 ( when f' is becoming zero but not changing their sign.)

8. f' has maximum value at x = 2 ( upper most vertex of graph)

9. f" has it maximum value at x= 1 ( inflection point of f'(x) )

10. Your answer is correct.

11, yes because f'(x) has slope from positive to negative and then positive ( -5, 1).

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