Mathematics: Post your doubts here!

[Thought blocker]

I still don't get it
the f inverse is root of(x-3) + 2

Where is the question ? I'll try to solve it.

 

[386423]

10 The function f is defined by f : x → 2x
2 − 12x + 13 for 0 ≤ x ≤ A, where A is a constant.
(i) Express f(x) in the form a(x + b)
2 + c, where a, b and c are constants. [3]
(ii) State the value of A for which the graph of y = f(x) has a line of symmetry. [1]
(iii) When A has this value, find the range of f. [2]
The function g is defined by g : x → 2x
2 − 12x + 13 for x ≥ 4.
(iv) Explain why g has an inverse. [1]
(v) Obtain an expression, in terms of x, for g−1
(x). [3]

 

[ZaqZainab]

Actually it would be done this way. I was also confused about it

f (x) = hg(x)
(x-2)^2 + 3
Now, since hg is a composite function, the value of g, which is x-2, would have to be put into the value of h in such a way that it becomes, (x-2)^2 + 3.

Thus h would be x^2 + 3. You can put in the value of g now and you will get f(x)

how did you end up with x^2 +3

 

[ZaqZainab]

Where is the question ? I'll try to solve it.

 

[386423]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

Assalamoalaikum!!

UPDATE: Link to Sequences Help by destined007 added!

10 The function f is defined by f : x → 2x
2 − 12x + 13 for 0 ≤ x ≤ A, where A is a constant.
(i) Express f(x) in the form a(x + b)
2 + c, where a, b and c are constants. [3]
(ii) State the value of A for which the graph of y = f(x) has a line of symmetry. [1]
(iii) When A has this value, find the range of f. [2]
The function g is defined by g : x → 2x
2 − 12x + 13 for x ≥ 4.
(iv) Explain why g has an inverse. [1]
(v) Obtain an expression, in terms of x, for g−1
(x). [3]
help please

 

[Thought blocker]

View attachment 39573

Which paper is it ?

 

[Thought blocker]

View attachment 39573

Is h = 2X - 5 ?

 

[ZaqZainab]

Is h = 2X - 5 ?

nope

 

[ZaqZainab]

Which paper is it ?

lol i don't remember it was a winter paper though

 

[Thought blocker]

nope

Then is it Root of X-3 +2 ?

 

[Thought blocker]

lol i don't remember it was a winter paper though

What is the answer ?

 

[ZaqZainab]

Then is it Root of X-3 +2 ?

Thats what i get using Suchal Riaz method

 

[ZaqZainab]

What is the answer ?

its x^2 + 3

 

[Thought blocker]

its x^2 + 3

Let me try again ! Till that you solve 386423 doubt

 

[usama321]

View attachment 39573

Its a composite function hg(x)

Since g(x) = x-2 and we have to make it equal to (x-2)^2 + 3,
Now in composite functions, we take the value of one function, and replace the variable x in the other function with that value.

Now for hg be equal to (x-2)^2 + 3, we need to do g^2 + 3. As in composite functions we replace the x with another function, we will put that x back in. Thus it would become x^2 + 3. Now put instead of x the value of g(x), and it becomes
(x-2)^2 + 3

 

[Thought blocker]

Its a composite function hg(x)

Since g(x) = x-2 and we have to make it equal to (x-2)^2 + 3,
Now in composite functions, we take the value of one function, and replace the variable x in the other function with that value.

Now for hg be equal to (x-2)^2 + 3, we need to do g^2 + 3. As in composite functions we replace the x with another function, we will put that x back in. Thus it would become x^2 + 3. Now put instead of x the value of g(x), and it becomes
(x-2)^2 + 3

Was it that easy ?

 

[usama321]

Was it that easy ?

Kindly tell me what you define as easy. If you take for granted that a one mark question is easy, than of course it was easy. If you define a question that requires more than average logic, and frustrates you as easy, than it was easy once again.

p.s i saw someone else's solution. wasn't able to do it myself

 

[Thought blocker]

Kindly tell me what you define as easy. If you take for granted that a one mark question is easy, than of course it was easy. If you define a question that requires more than average logic, and frustrates you as easy, than it was easy once again.

p.s i saw someone else's solution. wasn't able to do it myself

Oh, but I get to know your solution well, it was easy, I too got the same way! But I thought it might be wrong. SO I DIDNT POST IT!

 

[abruzzi]

Thanks a lot!
Could you please help me with part ii of the same question? Sorry for bothering..

 

[ZaqZainab]

Thanks a lot!
Could you please help me with part ii of the same question? Sorry for bothering..

which question?