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Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

An implicity function is one that is not expressed in the form y = f(x) such as the equation in the question. Instead of rearranging the equation to make y the subject, the equation can be differentiated using a technique called implicity differentiation. This involves differentiating each term on both sides of the equation. Differentiating x^3 will give 3x^2 and differentiating 34 will give 0. However differentiating y^4 will give (4y^3) X (dy/dx). This is achieved by using the chaing rule whereby d(y^4)/dx = (d(y^4)/dy) X (dy/dx).

This question is asking to find the values for x, such that h(x) is strictly less than g(x). We can write this as 9^x + 9 < 10*3^x and solve for x as follows.
9^x + 9 < 10*3^x
=> 3^(2x) - 10*3^x + 9 < 0
We let t = 3^x :
=> t^2 - 10*t + 9 < 0
=> (t - 9)(t - 1) < 0
By either sketching the quadratic, or by a sign diagram we find the values of t that satisfy this inequality :
1 < t < 9.
By substituting t = 3^x again, we find :
1 < 3^x < 9
=> 3^0 < 3^x < 3^2
=> 0 < x < 2.

Consider the Left-Hand-Side (LHS) of the equation first. LHS: 1/2 + 1 + 2 + 2^2 + ... + 2^10. We identify this as a geometric series by noticing that dividing any term u_(n+1) by the preceding term n the result is 2, eg. 1/(1/2) = 2. We also note that there are 12 terms, and that the first term is 1/2. From the formula booklet, section 1.1, we can find an equation for the sum of a finite geometric series: S_n = u_1(1-r^n)/(1-r). Where u_1 is the first term, r is the ratio of successive terms (u_(k+1))/u_k and n is the number of terms. In our case these take the values: u_1 = 1/2 , r = 2 , and n = 12. Substituting these in the equation we have : S_12 = 1/2(1 - 2^12)/(1 - 2)
= 1/2(2^12 - 1)
= 1*2^11 - 1/2
We can compare this last result for S_12 with the RHS of the original equation RHS: a*2^b + c, to find a = 1, b = 11, and c = -1/2.

Solution of all integration problems starts with investigation of a given function. We canâ€™t represent it with standard functions, i.e. linear, power or trigonometric. Second step would be to try substitution method, were we substitute f(x) with other function, letâ€™s say u(x). Have the same argument â€“ x, but are different. However, it is not some random function. We divide our initial function f(x) into u(x) and uâ€™(x), new function and new functionâ€™s derivative. Essentially, this is called u substitution method.
Now, if you look closely, f(x) consist from linear function (4x) and quadratic (x^2 + 1). Letâ€™s try u substitution, assuming that u(x)= x^2 + 1. Therefore, uâ€™(x) = 2x. 2x is not a 4x, but we could factor out 2, which gives us 2*2x, or 2 * uâ€™(x). Thus, f(x) = (2 * uâ€™(x)) / u(x) = 2 (uâ€™(x) / u(x)). uâ€™(x) is the same as du(x) / dx. 2 âˆ«(1 / u(x)) * (du(x) / dx) dx. We imagine as dx is divided by dx and we are left with du(x).
âˆ´ 2 âˆ«1 / u(x) du(x) = 2 * ln(u(x)) + C. Now, substituting u(x) = x^2 + 1 will give us: 2 * ln(x^2 + 1) + C. To find an area under a curve we should subtract integral with x = 0 from integral with x = 2. Area = (2 * ln(2^2 + 1) + C) â€“(2 * ln(0^2 + 1) + C) = 2 * ln(5) â€“ 2 * ln(1) = 2 * ln(5)
And we are done.

Answered by Igors D.

Studies Natural Science at York

Use quotient rule:
f'(x)=((x+1)2(x+2)+(x+2)^2)/((x+1)^2)
and simplify the above

This is a very common IB-level question,
which is solved with using the technique of a "disguised quadratic".
The first thing we want to do in finding the inverse is swapping the places of x and y in the equation y=f(x), to obtain x=f(y), because finding the inverse is equivalent to reflecting the curve in the line y=x, i.e. interchanging the x and y coordinates. Rearranging the equation x=f(y) to obtain y in terms of x will complete the task.
As such, we have: x=(e^y-e^(-y))/2. We want to solve for y to find the inverse. Firstly, multiply both sides by 2 to obtain: 2x=e^x-e^(-x). Now comes the tricky part: this is actually a quadratic in disguise! We can see this firstly from the rules of exponents: e^(-y)=1/(e^y), so we have: 2x=e^y-1/(e^y). Let us write e^y=p for simplicity: this will render it easier to spot the quadratic. Then 2x=p-1/p, so 2xp=p^2-1, which rearranges to p^2-2xp-1=0. It may not seem like we have done anything special, but remember that p=e^y, so if we can get p in terms of x, we will have e^y in terms of x, so taking a natural logarithm of the eventual expression will gives us the inverse.
Now we can use the quadratic formula to solve for p, where a=1, b=-2x, and c=-1.
We have: p= (2x +- sqrt((-2x)^2+4))/2, which is the same as p= (2x+-sqrt( 4x^2+4))/2. Now factor a 4 out of the expression in the square root to obtain: p= (2x+-sqrt( 4(x^2+1)))/2, or bringing the 4 outside the root: p= x+- sqrt( x^2+1).
Recall that p=e^y so we wouldn't want the expression for p to be negative, since we then wouldn't be able to take the logarithm of a negative value. As such, we take the positive square root in the expression for p: p=x + sqrt(x^2+1). So e^y= x + sqrt(x^2+1), or y= ln(x+sqrt(x^2+1)).
And there we have it: the inverse of y=f(x), where f(x)=(e^x-e^(-x))/2, is y= ln(x+sqrt(x^2+1)).

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