Transforming Functions: Solving g(x) = 2f(-x+(3/2))

[AAAA]

Homework Statement

If f(x)=|x-1/2|-5 determine g(x)=2f(-x+(3/2))

Homework Equations

The Attempt at a Solution

Well, I tried to factor out the k-value in the g(x) formula.
So I was left with:

g(x)=2f(-1)(x-3/2)

Then I multiply f(x) by 2 and am left with:
g(x)=2|x-(1/2)|-10

Then I subtract the 3/2 from 1/2 and am left with -2:
g(x)=2|x-2|-10

Then I apply the negative k-value and am left with
g(x)=2|-x+2|-10I checked on desmos, and that answer is wrong. It should be:
g(x)=2|-x+1|-10I've asked friends, looked online, in my textbook, and in my notes for things relating to this, and after 3 hours, came up empty-handed. If anyone could tell me where I went wrong. I would be very grateful. If you could also go step-by-step solving this problem, I would appreciate it.

Thanks.

 

[Ray Vickson]

Homework Statement

If f(x)=|x-1/2|-5 determine g(x)=2f(-x+(3/2))

Homework Equations

The Attempt at a Solution

Well, I tried to factor out the k-value in the g(x) formula.
So I was left with:

g(x)=2f(-1)(x-3/2)

Then I multiply f(x) by 2 and am left with:
g(x)=2|x-(1/2)|-10

Then I subtract the 3/2 from 1/2 and am left with -2:
g(x)=2|x-2|-10

Then I apply the negative k-value and am left with
g(x)=2|-x+2|-10I checked on desmos, and that answer is wrong. It should be:
g(x)=2|-x+1|-10I've asked friends, looked online, in my textbook, and in my notes for things relating to this, and after 3 hours, came up empty-handed. If anyone could tell me where I went wrong. I would be very grateful. If you could also go step-by-step solving this problem, I would appreciate it.

Thanks.

To get g(x), replace x everywhere (in the f(x) formula) by -x + (3/2); after that, multiply the whole thing by 2. In other words, $g(x) = 2 \left. f(t) \right|_{t = -x + 3/2}$.

 

[AAAA]

In other words, g(x)=2f(t)|t=−x+3/2g(x) = 2 \left. f(t) \right|_{t = -x + 3/2}.

I don't follow the last bit. I think the absolute value sign got messed up. Thanks for responding!

 

[AAAA]

I get it now! Thanks so much! Now I can finally move on

 

[Ray Vickson]

I don't follow the last bit. I think the absolute value sign got messed up. Thanks for responding!

Just to be clear: the notation $f(t)|_{t = u}$ does NOT mean there is a missing absolute-value sign. The notation is shorthand for "$f(t)$, evaluated at $t = u$". Of course, that is just $f(u)$, but since you seemed to be confused by notation such as $f(-x + 3/2)$ (that is, where $u = -x + 3/2$) I used the alternate notation instead. It is similar to the notation used in expressing definite integrals, such as
\int f(x) \, dx = F(x) \Rightarrow \int_a^b f(x) \, dx = F(x)|_{x=a}^{b} = F(b) - F(a).