Calculus Calculus Math Forum

 January 8th, 2014, 11:20 PM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Gradient function A curve with gradient function 4x-(4/(x^2)) has a turning point at (h,12). Find the value of h Find the equation of the curve
 January 9th, 2014, 12:10 AM #2 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Gradient function $y'(x) = 4x - 4x^{-2} \ \Rightarrow \ y(x) = 2x^2 + 4x^{-1} + k$ $y(h)= 12 \ \Rightarrow \ 2h^2 +\frac{4}{h} + k = 12 \ \cdots (1)$ $y'(h) = 0 \ \Rightarrow \ 4h - 4h^{-2} = 0 \ \cdots (2)$. Solving the above two equations simultaneously reveals all remaining info about your curve (namely the value of $k$).
 January 9th, 2014, 05:07 AM #3 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Re: Gradient function How to solve the simultaneous equation?
 January 9th, 2014, 06:23 AM #4 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Gradient function Equation (2) has only one possible real solution which is $h= 1$. Plug this value into equation (1) and then solve for $k$.
 January 9th, 2014, 06:29 AM #5 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Re: Gradient function Can you do the step-by step for solving the equation 2? I haven't learn to solve the equation which has a power to -2.
 January 9th, 2014, 10:55 AM #6 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Gradient function $4h - \frac{4}{h^2}= 0 \ \Rightarrow \ 4h^3 - 4 = 0 \ \Rightarrow \ h^3 = 1$. On the first implication, I just multiplied through by $h^2$ to clear the negative power, which is what you should always think about doing first in polynomial equations with negative powers. Observe that $h \neq 0$ (why?)

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post jiasyuen Calculus 1 January 9th, 2014 12:51 PM jiasyuen Calculus 3 January 9th, 2014 02:27 AM BenR Calculus 0 December 12th, 2012 01:33 AM Wooseok Nam Applied Math 0 May 20th, 2010 12:44 AM Mediocre1981 Calculus 3 August 29th, 2008 03:01 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top