further maths question

this is the solution given to us, so if you could help me with this method used.

i have not understood which parts of the graphs to take to do simultaneous

eg. she took for a,d y=x^2-9. y=1-2x the positive parts but why?

a is between f(x) and g(x) but d between -g(x) and f(x) so why take those 2

The argument in each modulus changes sign when it becomes zero. So the roots of x^2-9 are +/-3 and 1-2x is zero when x=1/2. So by considering each of the intervals -inf..-3, -3..1/2, 1/2..3 and 3..inf. you can remove the modulus signs by multiplyinng the arguments by +/-1 to make them correct and solve the resulting quadratics as usual.

So for -inf..-3, the lhs argument and the rhs argument are both positive so solve
x^2-9 = 1-2x
For -3..1/2, the lhs is negative, but the rhs is positive so solve
-x^2+9 = 1-2x
...

There are only really 2 different equations to solve as you can multiply through by -1, but 4 intervals, so each equation will refer to 2 intervals. Just think about the signs of the lhs and rhs. If theyre the same, use the first case (above). If theyre diferent, use the second case.

at school we use a different method, could you help me with the method i have explained i have also uploaded a picture so you could see the graphical method i was confused on which parts of the modulus graphs to take to do the simultaneous

tbh, thats pretty much the method on the picture as far as I can work out. The first part computes the roots of the two sides (to work out the intervals), then you just sketch it and put on the signs for f(x), the left hand side argument, and g(x), the right hand side argument. When the two signs are equal you solve
f(x) = g(x)
When the differ you solve
f(x) = -g(x)
So pretty much as I described?

you have described the method i had a different question on the parts of the graph you take anyway,
so you always do f(x) = g(x) f(x) = -g(x)?
then why do the graph then if its always like that?

Im not sure I understand your question 100%, but to try and elaborate a bit.

Your graph is of |f(x)| and |g(x)|. By definiton, this must be +/-f(x) and +/-g(x) with the sign chosen appropriately for the relevant interval to make sure the "output" is always positive (non negative).

On interval A, so -inf..-3, both f(x) and g(x) are positive. So |f(x)|=|g(x)| is equivalent to f(x)=g(x).
On interval B, so -3..1/2, f(x) is negative and g(x) is positive. So |f(x)|=|g(x)| is equivalent to -f(x)=g(x).
On interval C, so 1/2..3, both f(x) and g(x) are negative. So |f(x)|=|g(x)| is equivalent to -f(x)=-g(x).
On interval D, so 3..inf, f(x) is postive and g(x) is negative. So |f(x)|=|g(x)| is equivalent to f(x)=-g(x).

Intervals A and C produce the same equation f(x)=g(x) as As equation is simply Cs equation multplied by -1
Intervals B and D produce the same equation f(x)=-g(x) as Bs equation is simply D equation multiplied by -1

If this isnt what youre unsure about, just highlight which part.

yes thank you the intervals where the things i was not sure about, in this type of question where modulus is equal to modulus i can always do f(x) = g(x) f(x) = -g(x)?

Without being funny, you simply have to work it through pretty much as described in the original reply and the subsequent ones. When the argument of the modulus function is negative, the output is simply the argument multiplied by -1. So look for the roots (assuming they corresponds to a sign change) and they defines the intervals. In this case the problem is simply |f|=|g| so there will be two independent equations f=g and f=-g as the other two can be identity transformed to those. By solving the two equations and checking the solutions match the domain (union of the intervals), youre good.

An insightful sketch of f and g or |f| and |g| which defines the intervals and hence the sign of the function is a sensible way to approach such questions.

hi its me again sorry to bother you i was wondering if you could help me with part b of another question i attached, i found thederevative of part if that helpful

Im presuming your derivative is correct? Youre told the tangent at point x=a has a y intercept of 0, so form the tangent equation and see what you get?

isnt that pretty long and difficult towork with that what i got

Your derivative isnt correct its (...)^(1/2) so what happens to the power?

yes sorry its -1/2

So you have the equation of the tangent and it must be equal to the curve at that point.
Tbh, with a question like this, its sometimes worth working back from the answer a line or two and see if you can see the argument to finish it off.

could you explain the way we are going to prove that a is a solution first please

Ive tried to twice and the hint about working back from the solution. What do you get for the tangent at that point?

i dont really get what youre explaining
you mean the equation of the tangent?

yes what is the equation of the tangent to the curve at that point.