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• Level: GCSE
• Subject: Maths
• Word count: 3595

Extracts from this document...

Introduction

Middle

...etc. In each case the second bracket contains n terms. I can summarise this with the formula: an - bn = (a - b)(a(n-1) + a(n-2)b + a(n-3)b� +....+ b(n-1)) [With n terms in the second bracket] If I apply this rule to y = xn I can prove that my rule works for all values of n as a positive integer. Gradient = (x + h) - xn (x + h) - x = [(x + h) - x][(x + h) + (x + h) x +...+ x ] h = h[(x + h) + (x + h) x +...+ x ] (cancel x) h = [(x + h)(n-1) + (x + h)(n-2)x +...+ x(n-1)] (cancel h) If I now expand all the brackets and replace h with 0, the only terms I am left with are x(n-1). There will be n of these terms as there are always n terms in the second bracket. Gradient Function = x(n-1) + x(n-1) + x(n-1) +...+ x(n-1) [With n terms of x(n-1)] =nx(n-1) I have therefore proved that for all curves of the form y = xn, the formula nx(n-1) gives the gradient function of the curve for any value of n as a positive integer. Finding a Rule on Other Curves I now wish to see if I can find a rule for curves of the form y = Axn. I will again work through two examples of curves of this nature to attempt to find a rule which finds the gradient function of any of these curves. 1. y = 2x� Gradient = 2(x + h)� - 2x� (x + h) - x = 2(x� + 2xh + h�) - 2x� (expand brackets) h = 2x� + 4xh + 2h� - 2x� (expand brackets) h = 4xh + 2h� (cancel 2x�) h = 4x + 2h (cancel h) as h tends to 0 GF tends to 4x 2. ...read more.

Conclusion

Testing y = x-�; Hypothetical Gradient at Point x =0.5 is -16. Graph H shows that my rule also works for the graph y = x-� as my hypothetical gradient at the point x = 0.5 was the same as that which I calculated using the 'Small Increments Method'. Proving I will now try to prove my rule for graphs of the form y = x-n. My rule should lead to a formula of form GF = -nx-(n-1) y = x-n can be written as y = 1/xn Gradient = [1/(x + h) ] - [1/x ] (x + h) - x 1/{[(x + h) - x][(x + h) + (x + h) x +...+ x ]} h 1/{[h][(x + h) + (x + h) x +...+ x ]} h 1/[(x + h)(n-1) + (x + h)(n-2)x +...+ x(n-1)] (with n terms in the bracket) 1/n[(x + h)(n-1)] -nx-n-1 Fractional Powers I will now use my rule to hypothesise the gradient function for a graph of the form y = xp/q (where p and q are positive integers). I will then test it by using the 'Small Increments Method' to see if my rule works. I will then try to prove it using the 'Small Increments of Size "h" Method'. Testing y = x ; Hypothetical Gradient at Point x = 0.5 is 2. Graph I shows that my rule also works for the graph y = x1/2 as my hypothetical gradient at the point x = was the same as that which I calculated using the 'Small Increments Method'. Proving I will now try to prove my rule for graphs of the form y = xp/q. My rule should lead to a formula of form GF = (p/q)x[(p/q)-1]. Gradient = (x + h) - xp/q (x + h) - x = [(x + h) - x][(x + h) + (x + h) x +...+ x ] h = [h][(x + h) + (x + h) x +...+ x ] h = [(x + h)(p/q-1) + (x + h)(p/q -2)x +...+ x(p/q-1)] (with (p/q) terms in the bracket) = (p/q)x(p/q) ...read more.

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