# maths help (1 Viewer)

#### d3st1nyLiang

1. how do you integrate this from -2 to 2?
http://integrals.wolfram.com/index.jsp?expr=xe^x^k&random=false

2. How do you show that the line y = x+1 is tangent to y = e^k

and prove with a diagram the e^x-x-1 = 0 has only 1 solution

3. Solve e^t - e^-t = 2

#### jet

##### Banned
$\bg_white e^t - e^{-t} = 2 \\ e^t - \frac{1}{e^t} - 2 = 0 \\ \left (e^t \right )^2 - 2e^t - 1 =0 \\ \therefore e^t = \frac{2 \pm \sqrt{4 + 4}}{2} \\ e^t = \frac{2 \pm \sqrt{8}}{2} \\ = 1 \pm \sqrt{2} \\ \therefore t = ln \left (1 + \sqrt{2} \right ) \text{ since }1 - \sqrt{2}\text{ is negative}$

#### jet

##### Banned
For the second question, you find the points where the gradient of y = e^x is 1 (i.e the gradient of y = x + 1). This should give you a single point. Drawing y = e^x on a graph and showing the tangent, it is obvious that the tangent only touches y = e^x once. Hence, there is one solution to e^x = x + 1 or, rewritten, e^x - x - 1 = 0

#### jet

##### Banned
A question - are these 3 unit or 2 unit questions?

2unit?

#### jet

##### Banned
Well I don't see a method of doing that first question without 3u and 4u methods.