How do I solve this derivative question?

I have no idea how to make a graph like that.

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6 Answers

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  • 1 month ago

    f(x) = (3x+1) if x</= 1 or (-2x+6) if x >/= 1

    h(x) = f(g(x) 

    h(x) = (3(2x+1) + 1 if (2x+1)</= 1

    ==> h(x) = 6x + 4 if 2x </= 0

    or h(x) = (-2(2x+1) +6) if (2x+1)>/= 1

    ==> h(x) = (-4x +4) if 2x >/= 0

    if x = 1 then 2x> 0

    ==> h(x) = -4x +4

    ==> h'(x) = - 4

    ==> h'(-1) = -4

  • 1 month ago

    The rise slope and fall slope are different.

    E non existent.

  • ?
    Lv 7
    1 month ago

    Nonexistent .............ANS (Option E)

  • Dixon
    Lv 7
    1 month ago

    Since g(x) = 2x + 1

    f[g(x)] = f(2x + 1)

      

    Then

    h(x) = f(2x + 1)

      

    We are given f(x) on the graph, so we can transform it to f(2x + 1) as shown in the sketches below.

      

    Then by inspection, at x = 1 the gradient is the slope of the downward line segment.

    h'(1) = (y2 - y1)/(x2 - x1)

    h'(1) = -6/1.5

    h'(1) = -4 . . . . . . . . ANSWER

    We can also reason that if h'(1) does map onto the downward line segment, then adding 1 has no effect on the slope, but multiplying by 2 will make the slope twice as steep. And the original slope is -6/3 = -2, so that is OK.

    [Added: Some answers say there isn't a single value for the slope at x = 1 but the graph in the question is for f(x) not h(x) ]

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  • fooks
    Lv 7
    1 month ago

    1.

    Refer to the graph of the function f.

    The slopes of f(x) at x = 1 are different in the two line segments.

    When 0 < x ≤ 1: f(x) = m₁x + a

    When 1 ≤ x ≤ 4: f(x) = m₂x + b, where m₁ ≠ m₂

    When 0 < x ≤ 1:

    h(x) = f[g(x)]

    h(x) = f(2x + 1)

    h(x) = m₁(2x + 1) + a

    h(x) = 2m₁x + (m₁ + a)

    h'(x) = 2m₁

    h'(1) = 2m₁

    When 1 ≤ x ≤ 4:

    Similarly, h'(1) = 2m₂

    As the values of h'(1) are different (2m₁ ≠ 2m₂) in the two line segments, h'(1) is nonexistent.

    The answer: e. nonexistent.

  • david
    Lv 7
    1 month ago

    h'(1)  is nonexistent

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