Results 1 to 5 of 5

Math Help - tangent to curve

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    3

    Question tangent to curve

    can someone explain to me how this question is done?




    Find an equation for the line tangent to the graph of the given function at the indicated point.





    f(x) = x3 x2 at (0, 0)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2008
    Posts
    135

    Solution

    f(x) = x3 x2 at (0, 0)

    To find the value for f(x), substitute, the x value in the original equation.

    f(0) = 0^3 - 0^2
    f(0) = 0 - 0
    f(0) = 0

    Now, take the first derivative of the original f(x) equation to get the slope of the tangent line.

    Use the power rule, which states that:

    If f(x) = x^n, then f'(x) = nx^(n-1).

    So, f'(x) = 3*x^2 - 2*x

    To find the slope substitute the x value into f'(x):

    f'(0) = 3*0^2 - 2*0
    f'(0) = 0 - 0
    f'(0) = 0

    So now we can set up the equation of the tangent line:

    y = mx + b. Note, f'(x) = m and f(x) = y.

    So, 0 = 0*0 + b

    b = 0.

    So the equation of the tangent line of f(x) at (0,0) is written as

    y = 0

    I hope this is useful.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    3
    thank you, it was really helpful


    I have another one. it says

    Find an expression for dy/dx.






    y = 6 and u = 8x - 5
    u^2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2008
    Posts
    135

    Solution

    y = 6 and u = 8x - 5
    u^2


    Find expression of the form dy/dx


    Is this y = 6/(u^2)?

    First, rewrite y as: y = 6*(u^-2). It is the same as the original y equation.

    Now, we differentiate. The right side is with respect to y and the left side is with respect to u:

    dy = 6*-2*(u^-3)du.

    This was obtained using the power rule, which states that if y = x^n, then dy/dx = n*x^n-1

    So, we have dy = 6*-2*(u^-3)du. We're halfway done since we have to have it in terms of dy/dx.

    Since u is denoted above as u = 8x - 5, we should differentiate this equation. The left side is with respect to u and the right side is with respect to x:

    du = 8 dx.

    Now, substitute this into the dy equation above along with the u value to get:
    dy = 6*-2*((8x-5)^-3)*8dx

    Divide both sides by dx and simplify to yield:

    dy/dx = -96*(8x-5)^-3.

    It wouldn't be a bad idea to review the power rule and the chain rule since both are involved in this problem.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2008
    Posts
    3

    thanks for your help cuz am no good at calculus

    Am currently doing a practice sheet but cant do half of it, need some help please. Here goes


    Determine the continuity of the function at the given points.






    f(x) = -1, for x =-2
    0.5, for x not equal to -2
    at x = -2 and x = -3







    Answer the question.






    Is the function given by
    f(x) =x+4/ x^2 -9x+8
    continous at x = 1? Why or why not?






    Find the intervals on which the function is continuous.






    Is the function given by
    f(x) = 1/ x+1^2 +2

    continuous on ? Why or why not?






    Find a simplified difference quotient for the function.






    f(x) = 9^x3






    Find a simplified form of the difference quotient for the function.






    f(x) = square root of x-4





    Calculate the requested derivative from the given information.





    Given f(u) = u-1/u+1, g(x) = u = square root of x, find (f g)'( 4).














    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tangent on a curve
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 30th 2010, 05:32 AM
  2. Replies: 6
    Last Post: April 7th 2010, 02:34 PM
  3. tangent to curve
    Posted in the Calculus Forum
    Replies: 7
    Last Post: August 6th 2009, 05:19 PM
  4. Tangent Curve to a Curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 25th 2009, 12:03 AM
  5. Tangent to the curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2008, 04:25 AM

Search Tags


/mathhelpforum @mathhelpforum