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Math Help - 5 CONCEPTUAL problems I need help with, thank you.

  1. #1
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    5 CONCEPTUAL problems I need help with, thank you.

    These are not difficult subjects, but they are tricky problems. To prepare for a test our professor gave us some practice problems to make sure we understand the concepts. I got most of them, but here are a few I can't figure out.


    45) Consider the curve defined by -8x2 +5xy+y3 =-149. Write and equation for the line tangent to f at the point (4, -1) and using this line approximate f(4.2).



    • a) -0.4
    • b) 0.4
    • c) -0.373
    • d) 0.373
    • e) -0.6

    CONCEPT: linearization, (and implicit differentiation).

    39) If f(x) = x3 +x and g(x) is the inverse function of f(x), then g'(1) =

    • a) -0.5
    • b) 0.003
    • c) 0.077
    • d) 0.25
    • e) 0.417
    • THIS ONE I FIGURED OUT ON WOLFRAM ALPHA, BUT CANNOT DO ON MY TI-89 CALCULATOR

    37) An object traveling in a straight line has position s(t) at time t. If the initial position is s(0) = 3 and the velocity of the object is v(t) = (1+2t2 )(1/3) , what is the position of the object at time t=2?

    • a)7.312
    • b)5.933
    • c)2.933
    • d)8.312
    • e)24
    • THE VELOCITY FUNCTION IS EXACTLY HOW THEY HAVE IT. WITHOUT AN EXTRA t IT IS HARD TO INTEGRATE, BUT I THINK I'M SUPPOSE TO BE ABLE TO FIGURE IT OUT CONCEPTUALLY.

    18) Let g be a twice-differentiable function with g'(x) > 0 and g''(x) > 0 for all real numbers x, such that g(4) = 12 and g(5) = 18. Of the following, which is a possible value for g(6)?

    • a)15
    • b)18
    • c)21
    • d)24
    • e)27

    16) The function f is differentiable and has values as shown in the table below. Both f and f' are strictly increasing on the interval 0<=x<=5. Which of the following could be the value of f'(3)?

    TABLE:

    x = 2.5 | 2.8 | 3.0| 3.1

    f(x)= 31.25 | 39.2 | 45| 48.05

    • a)20
    • b)27.5
    • c)29
    • d)30
    • e)48.05

    THANK YOU THANK YOU THANK YOU, FOR ANY HELP!!!
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  2. #2
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    Re: 5 CONCEPTUAL problems I need help with, thank you.

    Quote Originally Posted by jimmeyers View Post
    These are not difficult subjects, but they are tricky problems. To prepare for a test our professor gave us some practice problems to make sure we understand the concepts. I got most of them, but here are a few I can't figure out.


    45) Consider the curve defined by -8x^2 +5xy+y^3 =-149. Write and equation for the line tangent to f at the point (4, -1) and using this line approximate f(4.2).

    a straight-forward implicit derivative ... what did you get for dy/dx ?

    • a) -0.4
    • b) 0.4
    • c) -0.373
    • d) 0.373
    • e) -0.6

    CONCEPT: linearization, (and implicit differentiation).

    39) If f(x) = x3 +x and g(x) is the inverse function of f(x), then g'(1) =

    • a) -0.5
    • b) 0.003
    • c) 0.077
    • d) 0.25
    • e) 0.417
    • THIS ONE I FIGURED OUT ON WOLFRAM ALPHA, BUT CANNOT DO ON MY TI-89 CALCULATOR


    if f and g are inverses, then f[g(x)] = x ... take the derivative of this equation w/r to x and determine a relationship between the derivatives of the inverse functions

    37) An object traveling in a straight line has position s(t) at time t. If the initial position is s(0) = 3 and the velocity of the object is v(t) = (1+2t2 )(1/3) , what is the position of the object at time t=2?

    fundamental theorem ... s(t_2) - s(t_1) = \int_{t_1}^{t_2} v(t) \, dt

    • a)7.312
    • b)5.933
    • c)2.933
    • d)8.312
    • e)24
    • THE VELOCITY FUNCTION IS EXACTLY HOW THEY HAVE IT. WITHOUT AN EXTRA t IT IS HARD TO INTEGRATE, BUT I THINK I'M SUPPOSE TO BE ABLE TO FIGURE IT OUT CONCEPTUALLY.



    18) Let g be a twice-differentiable function with g'(x) > 0 and g''(x) > 0 for all real numbers x, such that g(4) = 12 and g(5) = 18. Of the following, which is a possible value for g(6)?

    g'(x) > 0 and g''(x) > 0 ... what does this tell you about the rate of change of g ?

    • a)15
    • b)18
    • c)21
    • d)24
    • e)27


    16) The function f is differentiable and has values as shown in the table below. Both f and f' are strictly increasing on the interval 0<=x<=5. Which of the following could be the value of f'(3)?

    same concept as the previous question

    TABLE:

    x = 2.5 | 2.8 | 3.0| 3.1

    f(x)= 31.25 | 39.2 | 45| 48.05

    • a)20
    • b)27.5
    • c)29
    • d)30
    • e)48.05
    ...
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  3. #3
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    Re: 5 CONCEPTUAL problems I need help with, thank you.

    Thank you for the response!

    For #39 I am still a little lost. First, I wrote it incorrectly: f(x) should be (x^3 + x). But also, I understand that g'(x) = 1/f'(g(x)) and this would suggest it goes the other way, f'(x) = 1/g'(f(x)). If I take the derivative of f(x) I get 3x^2+1 set that equal to 1/g'(f(x)) or 1/g'(1) then flip, I get g'(1) = 1/(3x^2+1). If I put 1 in for x I get 1/4 or .25, which is one of the options but incorrect.

    For #45, when I use the ImpDif function on my calculator (TI-89) it gives me 16x/3y^2. When I did it by hand I got the correct derivative and it all worked out (A). I have tried it a couple times - do you have any experience with these calculators? (Or perhaps ImpDif is the wrong command.)

    For #37, thank you, I had only tried it as an indefinite - it didn't cross my mind to try it as a definite integral which makes me think I am missing some basic intuition with derivatives/integrals.

    For 16 and 18, I figured them out using what I know about the first and second derivatives, but I felt like that was more of a good guess. Especially with 16. I believe, and correct me if I'm wrong, that I can only find the secant lines but those should still be getting steeper - does this sound right? (That is, unless I take the limit and algebraically work out the actual definition of a derivative - is this what you would do?)

    Anyway, I really appreciate the help! Thank you!
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  4. #4
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    Re: 5 CONCEPTUAL problems I need help with, thank you.

    Quote Originally Posted by jimmeyers View Post
    For #39 I am still a little lost. First, I wrote it incorrectly: f(x) should be (x^3 + x). But also, I understand that g'(x) = 1/f'(g(x)) and this would suggest it goes the other way, f'(x) = 1/g'(f(x)). If I take the derivative of f(x) I get 3x^2+1 set that equal to 1/g'(f(x)) or 1/g'(1) then flip, I get g'(1) = 1/(3x^2+1). If I put 1 in for x I get 1/4 or .25, which is one of the options but incorrect.
    g'(1) = \frac{1}{f'[g(1)]}

    since f and g are inverses, f(a) = 1 , g(1) = a

    x^3+x = 1 at x \approx 0.682 so a = 0.682

    g'(1) = \frac{1}{f'(a)} = \frac{1}{3a^2+1} = 0.417

    For #45, when I use the ImpDif function on my calculator (TI-89) it gives me 16x/3y^2. When I did it by hand I got the correct derivative and it all worked out (A). I have tried it a couple times - do you have any experience with these calculators? (Or perhaps ImpDif is the wrong command.)
    I did this by hand (as you should) ...

    \frac{dy}{dx} = \frac{16x-5y}{5x+3y^2}

    link w/ instructions for doing it on an 89 ...

    Module 13 - Implicit Differentiation

    For 16 and 18, I figured them out using what I know about the first and second derivatives, but I felt like that was more of a good guess. Especially with 16. I believe, and correct me if I'm wrong, that I can only find the secant lines but those should still be getting steeper - does this sound right? (That is, unless I take the limit and algebraically work out the actual definition of a derivative - is this what you would do?)
    correct
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