Solve the given problem involving functions: f(x)=|ax-b| and y=f(x)+c

[chwala]

Homework Statement

Kindly see attached.

Relevant Equations

Understanding of functions

A bit confusing here; what i did,

Using gradient, we have,

$m=\dfrac{7-1}{0-3}=-2$

$y=-2x+7$

Since there is a Vertex, we have the other $m_2=2$

thus,
$y=mx+c$

$1=2×3 +c$

$... y=2x-5$

$a=2, b=-5, c=12$ or$a=-2, b=5, c=2$

 

[PeroK]

It says $a$ and $b$ are positive, which excludes both your solutions. I don't understand your working, in any case.

What can you say about the vertex of this graph?

 

[BvU]

Perhaps you could draw a graph of the function for some arbitrary positive values of a and b. Then see what 'vertex' and 'crosses y-axis at ... ' mean.

$\$

 

[chwala]

I will check this later...

 

[SammyS]

A bit confusing here; what i did,

Using gradient, we have,

$m=\dfrac{7-1}{0-3}=-2$

$y=-2x+7$

Since there is a Vertex, we have the other $m_2=2$

thus,
$y=mx+c$

Yes, to the left of the vertex the graph coincides with the line given by: $\ y=-2x+7$.

To the right of the vertex, the graph coincides with a line having a slope of $2$ . I suppose that's what you mean by
" we have the other $m_2=2$ "

How can you justify having $c$ in the following, but not $b$ also?

$y=mx+c$

 

[chwala]

Homework Statement: Kindly see attached.
Relevant Equations: Understanding of functions

View attachment 327774

A bit confusing here; what i did,

Using gradient, we have,

$m=\dfrac{7-1}{0-3}=-2$

$y=-2x+7$

Since there is a Vertex, we have the other $m_2=2$

thus,
$y=mx+c$

$1=2×3 +c$

$... y=2x-5$

$a=2, b=-5, c=12$ or$a=-2, b=5, c=2$

Ok i now have,

$y=f(x) + c$

$y=-ax+b+c=-2x+7$

$a=2$

$b+c=7$

We know that the vertex is at $(3,1)$ therefore,

$2x-b=1$

$b=5, ⇒c=2$

 

[chwala]

Allow me to post the continuation of the question here; part (b)

Part (b) (i)

Let

$y=g(x)$

$y=px-q$

$px=y+q$

$x=\dfrac {y+q}{p}$

$g^{-1}( x)= \left[\dfrac {x+q}{p}\right]$

The domain of $g^{-1}( x)$ is $x≥0$

I need help on part b.ii

I have,

$px-q=\left[\dfrac {x+q}{p}\right]$

$p^2x-pq=x+q$

...
$x=\dfrac{q(p+1)}{(p+1)(p-1)}$

$x=\dfrac{q}{p-1}$

$p≠1$

 

[PeroK]

Ok i now have,

$y=f(x) + c$

$y=-ax+b+c=-2x+7$

$a=2$

$b+c=7$

We know that the vertex is at $(3,1)$ therefore,

$2x-b=1$

$b=5, ⇒c=2$

This is all wrong!

 

[SammyS]

Ok i now have,

$y=f(x) + c$

$y=-ax+b+c=-2x+7$

$a=2$

$b+c=7$

We know that the vertex is at $(3,1)$ therefore,

$2x-b=1$

$b=5, ⇒c=2$

You still do not have part a) correct.

 

[pbuk]

You still do not have part a) correct.

Which you @chwala should have been able to see immediately by substituting back into the question: $| 2 \cdot 3 - 5 | + 2 \ne 1$.

After more than 2,000 posts here you still have not learned this most basic and essential step: why not?

 

[chwala]

Homework Statement: Kindly see attached.
Relevant Equations: Understanding of functions

View attachment 327774

A bit confusing here; what i did,

Using gradient, we have,

$m=\dfrac{7-1}{0-3}=-2$

$y=-2x+7$

Since there is a Vertex, we have the other $m_2=2$

thus,
$y=mx+c$

$1=2×3 +c$

$... y=2x-5$

$a=2, b=-5, c=12$ or$a=-2, b=5, c=2$

Perhaps you could draw a graph of the function for some arbitrary positive values of a and b. Then see what 'vertex' and 'crosses y-axis at ... ' mean.

$\$

I did that; looks like i am missing something here;

I have

If $-ax+b+c=-2x+7$

then i know that $a=2$

Now since the vertex is at $(3,1)$ then $\dfrac{b}{a}=3$

$\dfrac{b}{2}=3$

$b=6, c=1$ using $f(x)=2x-5$ as my line of reference..

 

[chwala]

Which you @chwala should have been able to see immediately by substituting back into the question: $| 2 \cdot 3 - 5 | + 2 \ne 1$.

After more than 2,000 posts here you still have not learned this most basic and essential step: why not?

Noted @pbuk , your input highly appreciated sir.

 

[Mark44]

Ok i now have,

$y=f(x) + c$
$y=-ax+b+c=-2x+7$
$a=2$
$b+c=7$

We know that the vertex is at $(3,1)$ therefore,
$2x-b=1$
$b=5, ⇒c=2$

No. a = 2 is correct, but your values for b and c are not.
You have y = |2x - 5| + 2
This equation is satisfied for the point (0, 7) since 7 = |2*0 - 5| + 2, but is it satisfied for the points (3, 1) and (6, 7)?

Before you post a solution, you should check your work to ensure that you actually have found a solution.

 

[chwala]

I did that; looks like i am missing something here;

I have

If $-ax+b+c=-2x+7$

then i know that $a=2$

Now since the vertex is at $(3,1)$ then $\dfrac{b}{a}=3$

$\dfrac{b}{2}=3$

$b=6, c=1$ using $f(x)=2x-5$ as my line of reference..

This is my latest result on the post...sorry for the confusion @Mark44 .... such questions can be a bit crazy!

 

[Mark44]

This is my latest result on the post...sorry for the confusion @Mark44 .... such questions can be a bit crazy!

I see now that you had the correct values of b = 6, c = 1 in an earlier post that I missed. Since the original question asked for the parameters a, b, and c in the equation y = |ax - b| + c, you should not have used b and c with different values in a different equation.

 

[SammyS]

I did that; looks like i am missing something here;

I have

If $-ax+b+c=-2x+7$

then i know that $a=2$

Now since the vertex is at $(3,1)$ then $\dfrac{b}{a}=3$

$\dfrac{b}{2}=3$

$b=6, c=1$ using $f(x)=2x-5$ as my line of reference..

This is almost correct.

Yes, the correct values are what you have ; $a=2,\,b=6,\, \text{and } c=1$.

However, nowhere is $f(x)=2x-5$ a reference line.

$f(x)=|\,2x-6\,|$, so that for $x<=3$ we have $f(x)=-2x+6$ and for $x>=3$ we have $f(x)=2x-6$

Perhaps you are thinking of $y=f(x)+c$.

It is true that for $x>=3$ we have that $y=f(x)+c=2x-6+1=2x-5$ .

More briefly, it's $y=2x-5$

 

[chwala]

Allow me to post the continuation of the question here; part (b)

View attachment 327779

Part (b) (i)

Let

$y=g(x)$

$y=px-q$

$px=y+q$

$x=\dfrac {y+q}{p}$

$g^{-1}( x)= \left[\dfrac {x+q}{p}\right]$

The domain of $g^{-1}( x)$ is $x≥0$

I need help on part b.ii

I have,

$px-q=\left[\dfrac {x+q}{p}\right]$

$p^2x-pq=x+q$

...
$x=\dfrac{q(p+1)}{(p+1)(p-1)}$

$x=\dfrac{q}{p-1}$

$p≠1$

just making a correction here:The domain of $g^{-1}( x)$ is $x≥1$

 

[SammyS]

just making a correction here:

The domain of $g^{-1}( x)$ is $x≥1$

No. Not correct. You are ignoring an essential relationship between the domain and range for each of a function and its inverse

in addition to that:

To help anyone reading this thread, you likely should have pointed out that you are now addressing part (b) of the exercise. Also, part (b) was not included in the OP, but was given in an attachment in post #7, the post you quote regarding your "correction". Here I re-post that attachment:

In part (b) we have a function, $g(x)$, which is closely related to $f(x)$ found in part (a).
$g(x)$ is defined as $\displaystyle g(x)=\left| px-q\right|$, where constants $p$ and $q$ are positive. Also, $g(x)$ is defined only for $x\ge\dfrac q p$.

What significance does $x\ge\dfrac q p$ have besides defining the domain of $g(x)$ ?

 

[chwala]

No. Not correct. You are ignoring an essential relationship between the domain and range for each of a function and its inverse

in addition to that:

To help anyone reading this thread, you likely should have pointed out that you are now addressing part (b) of the exercise. Also, part (b) was not included in the OP, but was given in an attachment in post #7, the post you quote regarding your "correction". Here I re-post that attachment:

View attachment 328157

In part (b) we have a function, $g(x)$, which is closely related to $f(x)$ found in part (a).
$g(x)$ is defined as $\displaystyle g(x)=\left| px-q\right|$, where constants $p$ and $q$ are positive. Also, $g(x)$ is defined only for $x\ge\dfrac q p$.

What significance does $x\ge\dfrac q p$ have besides defining the domain of $g(x)$ ?

Let me analyze this again, the Modulus is/was a bit confusing,

Now if we have,

$g(x)=|px-q|$ then the domain of $g(x)$ will be $x∈\mathbb{R}$ and the range will be $g(x)≥0$

Now it follows that the domain of $g^{-1}(x)$ will be $x≥0$ and the range will be $g^{-1}(x)∈\mathbb{R}$

 

[SammyS]

Let me analyze this again, the Modulus is/was a bit confusing,

Now if we have,

$g(x)=|px-q|$ then the domain of $g(x)$ will be $x∈\mathbb{R}$ and the range will be $g(x)≥0$

Now it follows that the domain of $g^{-1}(x)$ will be $x≥0$ and the range will be $g^{-1}(x)∈\mathbb{R}$

It is explicitly stated that the function, $g$, is defined for $x\ge\dfrac q p$ . This implies that the domain of function $g$ is the interval $\displaystyle \left[\dfrac q p,\, \infty \right)$ .

$p$ is positive so $x\in \left[\dfrac q p,\, \infty \right)$ implies that $px-q\ge0$.

However, if $px-q\ge0$, then we can drop the absolute value notation from the definition of function, $g$.