VCE Methods Question Thread!

[brightsky]

Hey all,
I have a question that I couldn't solve
Solve for x:
e^-x + 3= x
Thanks!
 

This is an example of what we call a transcendental equation. In general, transcendental equations cannot be solved algebraically. However, you can obtain quite a decent approximation using Newton's method. Newton's method is not on the Methods course, but if you're interested:

Let f(x) = e^(-x) - x + 3. We are required to solve f(x) = 0, which effectively means finding the x-intercept of the graph of y = f(x). First, we guess at an x-intercept. This initial guess does not need to be accurate. All we need is a random initial guess. Suppose that we guess x = 1. Now we note that f(1) does not equal to 0, but that doesn't matter. What we do now is construct the tangent to the curve at x = 1, and then find the x-intercept of the tangent. After a bit of number bashing, you'll find that the x-intercept of the tangent to the curve at x = 1 is x = 2.73105857863... We now let this be our next guess at the x-intercept of the graph of y = f(x). But again, we note that f(2.73105857863...) does not equal to 0, and so again, we construct the tangent to the curve at x = 2.73105857863..., and then find the x-intercept of the tangent, which happens to be x = 3.04471545135... Just as before, we let this be our next guess at the x-intercept of the graph of y = f(x). After you repeat this process a sufficient number of times, you'll find that your guess at the x-intercept of the graph of y = f(x) gets better and better. The actual x-intercept is x = 3.04747849102..., and so you can see that after repeating the process twice, we already have a pretty good approximation.

It's in the methods 3 & 4 Essential Cambridge book
There is a specific question where they ask you for a intersection concerning two graphs: e^-x +3 and -log_e(x+3)
I figured that if I made e^-x +3 =x, I'd find the answer the easier way. 

I'm assuming you mean -ln(x-3) rather than -ln(x+3)?

[SE_JM]

Hi

I've been trying to solve this for some time.. but I keep getting the answer wrong. Help

Make x the subject
y= 5*(1-e^-x)

thanks in advance!

[keltingmeith]

Hi

I've been trying to solve this for some time.. but I keep getting the answer wrong. Help

Make x the subject
y= 5*(1-e^-x)

thanks in advance!

We take it step by step, remembering to just slowly take everything away from the x:

[SE_JM]

We take it step by step, remembering to just slowly take everything away from the x:

Hi!
The answer says it's
x=log_e(5/(5-y))

is it the same thing?

thanks

[brightsky]

-ln(1-y/5)=-ln((5-y)/5)=ln(5/(5-y)).

[SE_JM]

-ln(1-y/5)=-ln((5-y)/5)=ln(5/(5-y)).

aha!
thanks for that~

[Chang Feng]

Regarding molodus functions:
What is the method/ process behind solving y=abs(x+2) + abs(x-3). May someone please explain, thanks.

[brightsky]

Consider three cases: x =< -2, -2 < x =< 3 and x > 3. For each case, simplify the expression on the RHS by removing the modulus signs in the appropriate way (leave as is if expression inside modulus is positive; multiply by -1 if negative), and solve the simplified equation for x. Reject any solutions that don't lie within the domain under consideration.

[keltingmeith]

Regarding molodus functions:
What is the method/ process behind solving y=abs(x+2) + abs(x-3). May someone please explain, thanks.

While brightsky is absolutely correct, I actually wrote a more in-depth explanation a few pages back here, which I thought you might be able to get some use out of.

[Chang Feng]

thanks for the reply brightsky, and EulerFan101 for the detailed solution = it helped a lot with the understanding (cause half times i feel like i'm purely applying formula, with limited knowledge of where formula came from).
I now also have another question that is about vectors, i'm still not quote sure what exactly are vector resolutes, and whats it purpose ( is it purely for finding the shortest distance- is the distance from a point to a line always the shortest if its perpendicular).
Thanks.

[Chang Feng]

and also regarding exponential and logarithmic functions:
why is the function y=1/2^x reflected in the y-axis, i'm still sort of confused. since with previous encountered functions such as x^2, if it was transformed to 1/2x^2 it wouldn't be reflected in y-axis. Could someone please explain why? and thanks for answering the questions.

[catherine_mimi]

Hey guys,
I have a question regarding the exponential graph

The question asks:
The number of people, N, who have a particular disease at time t years is given by
N=N₀e^kt
If the number initially is 20000 and the number decreases by 20% each year, find the value of k

I thought k= -0.2 (20/100) but the textbook says it's -0.223

Why? Shouldn't it exactly be -0.2?

[catherine_mimi]

and also regarding exponential and logarithmic functions:
why is the function y=1/2^x reflected in the y-axis, i'm still sort of confused. since with previous encountered functions such as x^2, if it was transformed to 1/2x^2 it wouldn't be reflected in y-axis. Could someone please explain why? and thanks for answering the questions.

I think it's because 1/2^x can be re-written as 2^(-x)
since the power of x is negative, it's reflected in the y-axis
is it clearer?
hope I made sense

[keltingmeith]

thanks for the reply brightsky, and EulerFan101 for the detailed solution = it helped a lot with the understanding (cause half times i feel like i'm purely applying formula, with limited knowledge of where formula came from).
I now also have another question that is about vectors, i'm still not quote sure what exactly are vector resolutes, and whats it purpose ( is it purely for finding the shortest distance- is the distance from a point to a line always the shortest if its perpendicular).
Thanks.

This is something best put into the specialist question thread, here. Just so you don't freak out anybody who's never heard of a "vector" before (particularly come exam time). (I'd answer it myself anyway, but I've gotta run, sorry. If you're not answered by the time I get back, I'll put something in)

[Zues]

dw got it