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Math Help - A couple of problems about derivitives

  1. #1
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    A couple of problems about derivitives

    The directions state, a. Find the difference quotient of f.
    b. f'(c) by comparing the limit of the difference quotient. Also, c is a constant.
    h=change in x.
    1.f(x)=f(x)=2-x^2 at c=0. the answer states: for a. -h and b. 0. For everytime I did the derivative formula, I get (2x-h). But when i plug 2 in for x and x^2 for h, it gets the right answer and that is what got me confused. Is it alwayz the case to use h when doing the derivative formula? If not, what are the exceptions?
    2. Find the standard form equation for the tangent line to y=f(x)at the specified point.
    f(x)= (x^2)-3x-5 where x=-2, the answer states:7x+y+9=0 I would know what rule to use if it had only an f(x) and a g(x). But on this one, I am not too sure.
    3. Find the coordinates of each point on the graph of the given function where the tangent is horizontal.
    f(x)=2x^3-7x^2+8x-3. the answer states: (1,0) and (4/3,-1/27). Same problem as #2 but not sure on the latter steps sense it is in coordinate form.
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  2. #2
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    Quote Originally Posted by driver327 View Post
    The directions state, a. Find the difference quotient of f.
    b. f'(c) by comparing the limit of the difference quotient. Also, c is a constant.
    h=change in x.
    1.f(x)=f(x)=2-x^2 at c=0. the answer states: for a. -h and b. 0. For everytime I did the derivative formula, I get (2x-h). But when i plug 2 in for x and x^2 for h, it gets the right answer and that is what got me confused. Is it alwayz the case to use h when doing the derivative formula? If not, what are the exceptions?
    2. Find the standard form equation for the tangent line to y=f(x)at the specified point.
    f(x)= (x^2)-3x-5 where x=-2, the answer states:7x+y+9=0 I would know what rule to use if it had only an f(x) and a g(x). But on this one, I am not too sure.
    3. Find the coordinates of each point on the graph of the given function where the tangent is horizontal.
    f(x)=2x^3-7x^2+8x-3. the answer states: (1,0) and (4/3,-1/27). Same problem as #2 but not sure on the latter steps sense it is in coordinate form.
    Hello,

    1. You have a function f with the equation y = f(x)
    2. You have a point P(a, f(a)) which is placed on the graph of f.
    3. Now the x-value changes a little bit. The new x-value is (a+h)
    4. You get another point on the graph Q(a+h, (f(a+h))
    5. Now you can calculate the slope between these 2 points:
    m = \frac{f(a+h) - f(a)}{(a+h)-a}
    6. This is the difference quotient for x = a.
    7. Because you can use this equation for every x of the domain of f normally the equation is written like this:
    m = \frac{f(x+h) - f(x)}{(x+h)-x} = \frac{f(x+h) - f(x)}{h}

    8. \lim_{h \rightarrow 0}(m) = f'(x)

    To #1:
    f(x) = 2-x^2

    m = \frac{2-(x+h)^2 - (2-x^2)}{h} = \frac{2-x^2 - 2hx-h^2 -2 +x^2 }{h}=\frac{-2hx-h^2}{h}=-2x-h . Difference quotient

    According to your problem x = 0 therefore m = -h

    f'(x) = \lim_{h \rightarrow 0}(-2x-h) = -2x . First derivation.

    According to your problem x = 0 therefore f'(0) = 0

    To #2:
    f(x) = x^2-3x-5 then f'(x) = 2x-3

    If x = -2 the f(-2) = 5 that means the point P(-2, 5) lies on the graph of f.
    The gradient of the graph at x = -2 must be the same as the slope of the tangent in P.
    m_{tangent} = f'(-2) = 2(-2)-3 = -7

    Now use point-slope-formula of a straight line to get the equation of the tangent:
    (y-y_1) = m \cdot (x-x_1) Plug in all known values into this equation and solve for y:

    (y-5) = (-7) \cdot (x-(-2)) \Longleftrightarrow y = -7x-14+5 \Longleftrightarrow y = -7x-9

    To#3:
    A horizontal line is a parallel to the x-axis that means it has the slope zero:
    The slope of the tangent is the same as the gradient of the function:

    f(x) = 2x^3-7x^2+8x-3 \Longrightarrow f'(x) = 6x^2 - 14x + 8

    Now f'(x) = 0. Therefore:

    6x^2-14x+8 = 0 \Longleftrightarrow (3x-4)(2x-2)=0 \Longleftrightarrow x = \frac{4}{3} \vee x = 1

    Plug in these x-values into the equation of the function to get the second coordinate of the tangent points.
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