x + h – x h
h (h³ + 4x³ +6x²h +4xh²) = h³ + 4x³ +6x²h +4xh²
h
H tends to 0 again here, and every term contains an h except for (4x³). Therefore, this is the gradient function for x4, and verifies that my earlier observations in the table method were correct.
y = x5
The general pattern here between the two values of x and the gradient, the value of the gradient function here is 5x4, as seen when comparing the two smaller tables above. Using this formula, the next 5 values of x and the gradient are:
General proof –
(x+h)5 – x5 = x^5 + 5x4h + 10x³h² + 10x²h² + 5xh4 +h5 – x5
x + h – x h
= h (5x4 +10x³h + 10x²h + 5xh³ + h4) = 5x4 +10x³h + 10x²h + 5xh³ + h4
h
Every term apart from 5x4 contains an h term, leaving every term to tend to 0. Therefore the gradient function of x5 = 5x4, which corresponds to my results earlier in the tables, proving my predictions of the gradient function was correct.
After investigating the gradient function where a is a constant 1, and n is always a whole number, I have concluded this section with a table of results for each value of n investigated so far –
Perhaps here I shall attempt to find a pattern on these data. By trying to find a general trend from this, I can conclude that when a, in axn, is constantly 1, and the values of n are whole numbers, the gradient function for this in general is:
nxn-1
However, I will need to prove this using binomial expansion. In this investigation, in the last section, for each value of n I have attempted to prove each gradient function is valid using general proof. Here, I shall try and do the same for when x = n.
As a guideline, I shall use the previous proof of x² along the way to compare to it to x^n.
x² =
Gradient of PQ = (x+h) ² - x² = x² + 2hx + h² - x² = 2hx + h² = 2x+h = 2x
x +h-x h h
x^n =
Gradient of PQ = (x+h)n – xn = xn + (nx n-1) (h) + n (n-1) (xn-2) h²
2!
+ n (n-1)(n-2) (x^n-3)(h)³ + n(n-1)(n-2)(n-3) (xn-4)(h4) + …hn
3! 4!
= (nx^n-1)(h) + n(n-1) (x^n-2)(h²) + … +h^n
2!
Therefore, (x+h)ⁿ - xⁿ = nxn-1 + n(n-1) (xn-2)(h) + … + hn-1
h 2!
Since h limits to 0, all values containing an h term are therefore all worth 0, so therefore the gradient function for xn = nxn-1
However, in this section, this is perhaps only true for positive integers for the value of n. It is clear enough that this works for all the values of n previously proven, but now I will attempt to try and prove this for fractional and negative powers to prove that this formula is correct. If there is irrefutable evidence to support this gradient function for xn, I will investigate different values of “a” in axn-1.
First I shall investigate fractional powers - I will try to investigate 2 or 3 of these to find a general rule.
y = x1/2
So far, I cannot conclude anything from these data since the numbers are not whole and therefore hard to work with. However, this evidence will help support the formula which I will attempt to create by using binomial expansion. I shall also see whether the gradient function I shall come up with corresponds to nx n-1.
If nx^n-1 is correct for all positive values of n, even non integers, I predict that the gradient function for x 1/2 is 1/2x-0.5. Here, it is a little hard to prove this in the normal sense as (x+h) is square rooted, so I will need to rationalize the numerator.
√ (x+h) – √x = √ (x+h) – √x
x + h -x h
√ (x+h) – √x x (√ (x+h) + √x) = (x+h) - x
h (√ (x+h) + √x) h (√(x+h) + √x)
I have multiplied this fraction to rationalize it, because (a-b) (a+b) = a² + b²
= h = 1 = 1
h(√(x+h) + √x) (√(x+h) + √x) √(x+0 + √x
= 1 = 1/2x^-1/2
2√x
This proves that my prediction was correct, and also fits with the formula nx^n-1. I will now compare the table method for √x with the gradient function 1/2x^-1/2:
1/2*1-0.5 = 0.5
1/2*2-0.5 = 0.3535
1/2*3-0.5 = 0.0.2887
1/2*4-0.5 = 0.25
These are exactly the same as the gradient; therefore this proves that I was correct. I shall now try and do the same for when n = ¼, to prove that nx^n-1 works for fractional powers, and not just when x = ½.
y=x¼
Like before, these data are very hard to work with and I cannot find a general pattern by looking at the gradients – they are not whole numbers. This will again, however, support the gradient function I will come up with using binomial expansion.
Therefore, this proves my prediction was correct, and works with the rule nx n-1. I will now test this formula with each relevant datum in the table.
These are exactly the same. Therefore this proves that all positive values of n in the equation nx n-1 apply to this formula. After I try to prove negative powers are correct, I shall thoroughly investigate, in depth, the overall gradient function for the curves y =ax n-1, where “a” is not constant.
But for now, however, I shall investigate negative values of n. The value of n I will choose is -1, or y= 1/x.
y = x-1
I predict that the gradient function for when n=-1 = -x -2, in correspondence to the formula nx n-1. I can tell whether these data do in fact agree with this gradient function, but I shall binomially expand this to verify this is the case:
1 . x - x+h . 1 = x - (x + h)
x + h x x+h x x(x+h) x(x+h)
x + h - x h
= x – (x+h) = -h = -1 = -1 = -1x-2.
hx(x+h) hx(x+h) x(x+0)* x2
*h tends to 0.
Voila, this gradient function is in accordance with nx n-1. By getting this far, I have nearly convinced myself that all values work, whether they are integers or not, or positive or not. However, thus far I have not perhaps produced enough evidence to support this theory, and therefore I shall attempt to prove one last n value in this section, where n=-2. Since I am nearly convinced that this formula is proven for any value, where “a” is a constant 1, I will just binomially prove y = x -2.
In accordance to the formula nx n-1, I predict that the gradient function for this = -2x-3.
y=x-2.
1 . x² - (x+h)² . 1* = x² - (x²+2hx+h²)**
(x + h) ² x² (x+h)² x² x²(x²+2hx+h²) x²(x²+2hx+h²)
x + h - x h
= x² - (x²+2hx+h²) = -2hx – h² = -2x – h = -2 – h/x = -2 – 0/x
hx²(x²+2hx+h²) hx²(x²+2hx+h²) x²(x²+2hx+h²) x(x²+2hx+h²) x(x²+((2)(0x))+0²)
= -2 = -2x-3 - here we assume x, is not 0.
x-3
* IN all the past equations I have rationalized the numerator to make them easier to work with and thus solve. Otherwise, they are thus irrational and impossible to solve, with all the knowledge of this topic I possess.
** h tends to 0.
I now move onto the second and final section of this piece of coursework – where we now change a new variable in this investigation - the value of a. I will try to combine this with different value of n, to see how it affects the end result.
axn
In this section, I will change the values of a by an increase of 1 in each example, and furthermore do the same for the value of n. While it is bad practice to change 2 variables at once as a rule, I have found the effect the changing n in the previous investigation, so is therefore valid. Firstly, I shall try to investigate 2x², and if successful, move onto greater values of a.
y=2x²
The gradient here is equal to 4x – I can see this by investigating the values of x, as compared to the gradient. I predict that the gradient for 5 will be 20. This can be shown here:
My hypothesis is correct. It will be also be useful to prove this binomially, however:
2(x+h) ² - 2x² = 2(x²+h²+2hx) – 2x² = 2x²+2h²+4hx -2x² = 2h²+4hx
x + h – x h h h
2h + 4x = ((2)(0)) +4x = 4x
Therefore the gradient function is correct. Now, I will change the value of n by 1 to see whether the same rule applies with different values of “a” and n.
y = 2x3
After looking through each value, I have observed a pattern. I have noticed that the gradient is always equal to 6x². I predict that when x = 5, the gradient will be equal to 150.
This has proved my hypothesis correct. Now, again I shall prove this binomially:
2(x+h) ³ - 2x³ = 2(x³ +3x²h + 3h²x+h³) – 2x³ = 2x³ + 6x²h + 6h²x +h³ - 2x³
x + h – x h h
= 6x²h + 6h²x +h³ = 6x² + 6hx + h² = 6x² + (6)(0x) + 0² = 6x²
h
I need not explain this any further. I will now investigate 2x4, and then try to find a general pattern for all the gradients of where a=2.
After looking through each value, I have observed a pattern. I have noticed that the gradient is always equal to 8x³. I predict that when x = 5, the gradient will be equal to 1000.
My theory was correct. I shall now try and binomially prove this…
2(x+h)4 – 2x4 = 2(x4 + h4 +4hx³ + 6x²h² + 4xh³) - 2x4 =
x + h – x h
2x4 + 2h4 +8hx³ + 12x²h² + 8xh³ - 2x4 = 2h4 +8hx³ + 12x²h² + 8xh³
h h
= 2h³ + 8x³ + 12x²h + 8xh³
After doing this so many times, I can tell that every term containing an h will disappear. The one exception to this = 8x³. This is the answer I had previously predicted, and therefore my theory was correct.
I have observed a pattern using these results. I have noticed that when a =2, if you multiply this through by the power, and subtract 1 from the power, you get the gradient function.
In other words = 2(nxn-1). This may apply to every value of a, thus giving us the overall gradient function, but I must find far more evidence than one solitary a value to give it any meaning.
If this is the case, however, I think that when the co-efficient is 3, the whole gradient will be 3(nxn-1). I shall try and investigate this.
3x²
After looking through each value, I have observed a pattern. I have noticed that the gradient is always equal to 6x. I predict that when x = 5, the gradient will be equal to 30.
The gradient here is 30, so therefore I was correct. Now, to the binomial proof…
3(x+h) ² - 3x² = 3(x² + 2hx + h²) - 3x² = 3x² + 6hx + 3h² - 3x²= 6hx + 3h²
x +h-x h h h
= 6x + 3h = 6x
Here, I have noticed a trend in these data. The gradient applies to nx n-1, because the power is always decreased by 1. Once multiplied through by the n value, and then n is decreased by 1, the answer is given. However, these may be isolated results and I shall have to clearly investigate other values of “a” and in the equation, to prove this is the case for all.
Here, I shall investigate some further “a” and x values to find the overall gradient function for axn; this will bring my investigation to a close. 3x³ is suitable for the next investigative values.
3x³
After looking through each value, I have observed a pattern. I have noticed that the gradient is always equal to 9x². I predict that when x = 5, the gradient will be equal to 225.
My theory was correct. Binomially expanding this, yet again, will allow me to get to the root of the equation, and see whether it is correct using algebra.
3(x+h)³ - x³ = 3(x³ +3x²h + 3h²x+h³) -3x³ = 3h(3x² +3hx + h²) =
x + h – x h h
3(3x² +3hx + h²) = 9x² + 9hx +3h² = 9x²* - “h” limits to 0.
This pattern is fairly apparent – with the same n value, only the “a” values change when investigated. I have, again, noticed that this follows a same general pattern as the previous table full of x² values.
The general pattern here is that 3(nxn-1) is the general formula to find out the gradient function, where a = 3. The next investigation of 3x4 will help me to find a pattern in general.
3x4
It is fairly obvious here, after looking through past x^4 results, where “a” is variable, that the gradient function here is equal to 12x³. Using these data, I predict that the next table containing the gradient of will display 1500.
Binomial proof shall verify this for us, though realistically enough, the table provides enough evidence.
3 (x+h) 4 – 3x4 = 3(x4 + h4 +4hx³ + 6x²h² + 4xh³) - 3x4 =
x + h – x h
(3h4 +12hx³ + 18x²h² + 12xh³) = 3h³ + 12x³ + 18x²h + 12xh² = 12x³
h
Here, h is too insignificant to bother multiplying through by, and every value contains an h term, apart from the gradient function, which is 12x³.
This fits in with a pattern i.e.
1(nx n-1) = gradient function, where a = 1
2(nx n-1) = gradient function, where a =2.
3(nxn-1) = gradient function, where a =3.
It is clear from these results that the overall equation for curves of ax^n is nax^n-1. I shall now test this using a value where “a” is a negative integer. A good candidate value of n for this next investigation would be -2x2.
-2x2
Using the formula naxn-1, I can see that the formula for -2x² is -4x, without doing any algebraic proof or tables. However, this is not necessarily the case, and therefore I will try to test it. I predict that for 5, the gradient will be -20.
It is correct, and therefore the formula is valid. Binomial expansion will show this:
-2(x+h) ² - -2x² = -2x² + -4hx + -2h²) --2 x² = -4hx -2h² = -4x -2h = -4x.
x +h-x h h
This is correct; therefore I am fully convinced that nax^n-1 is no doubt the overall gradient function for axn.
Charts of graphs I have done overall for curves of y = axn
“Odd functions”
It is clear to see that the overall equation for the gradient function of this, or any curve y=axn is naxn-1. I can observe this because when the values of the curves are substituted into this equation and then simplified the result is the gradient function of the particular curve.
Now it is very clear that this formula is the case, I will try to prove this, one last time, using binomial expansion.
The actual proof for axn is as follows –
axn + hanxn-1 + a n(n-1) (xn-2)h² + … + ahn – xn
2
h
I will try to prove this –
xn + (nhxn-1) + n(n-1) (xn-2)h² + … + hn – xn
2
h
Therefore the equation to work out the gradient formula for y=ax^n, will need to be multiplied by “a”, and therefore –
a (xn + (nhxn-1) + n(n-1) (xn-2)h² + … + hn – xn)
2
h
(axn + (anhxn-1) + an(n-1) (xn-2)h² + … + ahn – axn)
2
h
I can now multiply this through by h to get –
anxn-1 + an(n-1) (xn-2)h + … ahn-1.
2
Like all previous binomial expansion, every single value contains an h term, and continually limits to 0. The only term left over is
naxn-1
My investigation is now at end, and perhaps the coursework is now complete. The gradient function can find out any function of any curve of the y = axn.
“A loose end”
Some parts of this investigation require that I do a straight line graph for y = axn. I will only binomially expand this, and compare the result to the gradient function –
y= x
(x+h)-x = h = 1 =x0.
(x+h) – x h
If the formula for the gradient function naxn-1, 1(1x)1-1 = x0.
A straight line graph proof.
“An even greater loose end”
In this fairly brief section, I shall investigate what will happen when I try to add different values of axn, where “ a “ and n are variable. I will try to investigate two cases of where this happens. In the first case, I shall investigate using algebraic proof and tables to find the overall gradient function for axn + lxm.
2x² + 3x5
2(x+h)² -2x² + 3(x+h)5 – 3x5 =
(x+h) – x (x+h) – x
2(x²+ 2hx + h²) – 2x² + 3(x5 + 5x4h + 10x³h² + 10x²h² + 5xh4 +h5) – 3x5
h h
= 2x² + 4hx + 2h² - 2x² + 3x^5 + 15x4(h) + 30x³h² + 30x²h² + 15xh4 +h5 – 3x5
h h
= (4x + 2h) + (15x4 + 30x³h + 30x²h + 15xh3 +h4) = 4x + 15x4
= x (4 + 15x^3)
This seems to follow the same pattern as the previous investigation; each term follows the correct pattern in accordance with naxn-1 (2x² and 3x5 have the respective gradient functions of 4x and 15x4). I will need to try this, however, with 2 more cases, to find a respective pattern between all of them.
With the next method, I shall use solely the table/increment method to find respective gradients. It is perhaps not a preferred method to algebraic proof. However, I shall need to do this to see whether it is a better method.
3x3 + 2x3
Gradient + Gradient =
Gradient 3x3 + Gradient 2x3
From these results, I can figure out that the overall gradient function of each value added together is :
9x2 + 6x2 = x(9x + 6x)
This proves quite a clear pattern. Now it is clear that both functions correspond to the gradient function nax^n-1. However, the formula
naxn-1 + naxn-1 is not valid : this is because we are assuming that the a and x values are not equal to each other. Therefore, I shall use a different letter to use in these equations. I shall substitute “n” with “m” and “a” with “l”. Therefore the formula for adding powers together is :
nxn + lxm = naxn-1 + lmxm-1.
This is the overall formula for working out the basic functions of any curve of axn + lxm. If the powers are being subtracted from each other, the + is simply replaced by a -.
Now I have proved this, the coursework is complete.
