Could you check my work on these problems? Thanks!

[Steph Annie]

Hi! Could you check my work on these problems. My answers are in purple. Let me know if you have suggestions for changes or eliminations. Thanks!

Given the functions f(x) = 1/(x-2) and g(x) = log (x+1) with maximal domains:

a) What are the domains of f and g?

"f is not defined when the denominator equals zero. When x =2 then x-2 =0. So the domain of f is all real numbers except 2. x+1 > 0 or x > -1.

b) What is (f composition g)(x)? Also find the domain of f composition g.

(f composition g)(x) = 1/log(x+1)-2, so x>-1. Log (x+1)-2 must not equal zero.
Log(x+1)-2=0, log(x+1)=2, 10²=x+1, x=99. The domain of f composition g (x) is all real numbers greater than -1 except x =99.

c) What is (g composition f) (x)? ALso find the domain of g composition f.

log[1/(x-1)+1 = log[1+x-2/x-2] = log x-1/x-2. x can't be zero; logarithm must be positive, so (x-1)/(x-2) >0. The domain is all real numbers such that x<1 or x>2.

d) Is g one-to-one? If not, explain why not. If it is, find the inverse of g^-1(x) and give the range of g^-1.

Yes, g is one-to-one.
x= log (y+1)
10^x=y+1
y=10^x-1
The domain is all real numbers.

 

[wjm11]

Hi! Could you check my work on these problems. My answers are in purple. Let me know if you have suggestions for changes or eliminations. Thanks!

Given the functions f(x) = 1/(x-2) and g(x) = log (x+1) with maximal domains:

a) What are the domains of f and g?

"f is not defined when the denominator equals zero. When x =2 then x-2 =0. So the domain of f is all real numbers except 2. x+1 > 0 or x > -1.

b) What is (f composition g)(x)? Also find the domain of f composition g.

(f composition g)(x) = 1/log(x+1)-2, so x>-1. Log (x+1)-2 must not equal zero.
Log(x+1)-2=0, log(x+1)=2, 10²=x+1, x=99. The domain of f composition g (x) is all real numbers greater than -1 except x =99.

c) What is (g composition f) (x)? ALso find the domain of g composition f.

log[1/(x-1)+1 = log[1+x-2/x-2] = log x-1/x-2. x can't be zero; logarithm must be positive, so (x-1)/(x-2) >0. The domain is all real numbers such that x<1 or x>2.

d) Is g one-to-one? If not, explain why not. If it is, find the inverse of g^-1(x) and give the range of g^-1.

Yes, g is one-to-one.
x= log (y+1)
10^x=y+1
y=10^x-1
The domain is all real numbers.

Looks good to me. Nice work.

BTW, it is often very informative to graph these functions to see domain and range after calculating them.

 

[lookagain]

Hi! Could you check my work on these problems. My answers are in purple.
Let me know if you have suggestions for changes or eliminations. Thanks!

You have several errors, many of which are from not having required grouping symbols.


Given the functions f(x) = 1/(x-2) and g(x) = log (x+1) with maximal domains:

a) What are the domains of f and g?

"f is not defined when the denominator equals zero. When x =2 then x-2 =0. So the domain of f is all real numbers except 2.

x+1 > 0 or x > -1.* You never came out and stated explicitly in words that g is not defined when its argument is not positive.

I give you credit for showing these two equivalent inequalities.* Later on, you might explicitly state:
"So the domain of g is all real numbers greater than -1."


b) What is (f composition g)(x)? Also find the domain of f composition g.

(f composition g)(x) = 1/log(x+1)-2, so x>-1. Log (x+1)-2 must not equal zero. \(\displaystyle \ \ \ \)1/[log(x + 1) - 2] <---- grouping symbols required

Log(x+1)-2=0, log(x+1)=2, 10²=x+1, x=99. The domain of f composition g (x) is all real numbers greater than -1 except x = 99.


c) What is (g composition f) (x)? ALso find the domain of g composition f.


log[1/(x-1)+1 \(\displaystyle \ \ \ \ \)-------------> log[1/(x - 2) + 1]


= log[1+x-2/x-2] \(\displaystyle \ \ \ \ \)-----------> log[(1 + x - 2)/(x - 2)]


= log x-1/x-2.\(\displaystyle \ \ \ \ \) --------------> log[(x - 1)/(x - 2)]


x can't be zero; \(\displaystyle \ \ \ \ \) ------------> x can't be equal to 2.


logarithm must be positive, \(\displaystyle \ \ \ \ \)---------> The argument must be positive. The logarithm of the number may be negative, zero, or positive.


so (x-1)/(x-2) >0. \(\displaystyle \ \ \ \ \) ---------> Yes.


The domain is all real numbers such that x<1 or x>2. \(\displaystyle \ \ \ \ \)---------> Yes.


d) Is g one-to-one? If not, explain why not.

If it is, find the inverse of g^-1(x) \(\displaystyle \ \ \ \ \)This statement in the problem is worded wrong. It amounts to finding the
inverse of an inverse.

What it should have stated is closer to the following: "If it is, find the inverse, that being g^-1(x)."

and give the range of g^-1.

Yes, g is one-to-one.

x = log(y+1)

10^x = y+1

y = 10^x - 1 \(\displaystyle \ \ \ \ \)That is not the final form for the inverse. You want: g^-1(x) = 10^x - 1.


The domain is all real numbers. \(\displaystyle \ \ \ \ \)You didn't answer the question. It wanted the range of the
inverse of g(x). Hint: Without regard to mentioning x and y as a portion of them, respectively, the domain
of the original function is the same as the range of the inverse of the function.

.

 

[wjm11]

Thank you, lookagain, for your more rigorous/complete response to the OP. I (lazily) settled for a check of overall comprehension and interpreted the typos and missing grouping symbols as I believed they were intended.

To Steph Annie: please pay close attention to all the things lookagain took the time to point out.

 

[Steph Annie]

Thank you, lookagain, for your more rigorous/complete response to the OP. I (lazily) settled for a check of overall comprehension and interpreted the typos and missing grouping symbols as I believed they were intended.

To Steph Annie: please pay close attention to all the things lookagain took the time to point out.

Thank you both. I greatly appreciate it.

And there was one last question relating to #2. "Find all intercepts, both x-intercepts and y-intercepts of the following functions. Express each intercept as an ordered pair, for example, (0, -2)." I hadn't done the problem yet, so I'm now including it.

Part c.) h(x) = x³-x²-7x-5

I answered: y-intercept: y = -5, when x is equal to zero, so the y-intercept is -5.
x-intercept: 0 = x³-x²-7x-5
5 = x(x²-x-7)
5 = x(x-7)(x+1)
The x-intercept is approximately equal to 7.09

 

[Deleted member 4993]

Thank you both. I greatly appreciate it.

And there was one last question relating to #2. "Find all intercepts, both x-intercepts and y-intercepts of the following functions. Express each intercept as an ordered pair, for example, (0, -2)." I hadn't done the problem yet, so I'm now including it.

Part c.) h(x) = x³-x²-7x-5

I answered: y-intercept: y = -5, when x is equal to zero, so the y-intercept is -5.
x-intercept: 0 = x³-x²-7x-5
5 = x(x²-x-7)
5 = x(x-7)(x+1)
The x-intercept is approximately equal to 7.09

By observation, there is one x-intercept at x = -1

then

x³-x²-7x-5 = (x+1)(x² - 2x - 5) ... now find the rest of the intercepts

 

[wjm11]

Thank you both. I greatly appreciate it.

And there was one last question relating to #2. "Find all intercepts, both x-intercepts and y-intercepts of the following functions. Express each intercept as an ordered pair, for example, (0, -2)." I hadn't done the problem yet, so I'm now including it.

Part c.) h(x) = x³-x²-7x-5

I answered: y-intercept: y = -5, when x is equal to zero, so the y-intercept is -5.
x-intercept: 0 = x³-x²-7x-5
5 = x(x²-x-7)
5 = x(x-7)(x+1)
The x-intercept is approximately equal to 7.09

For the y intercept, you need to state the answer as an ordered pair: (0,-5)

I'm not sure what you are attempting to do after that. Moving the 5 over and then factoring is not correct. You need to have zero on one side and then factor everything else on the other side if you want to find the zeroes.

Do you know how to graph cubic equations? Do you know how to find the roots of cubic equations? Do you know how to attempt to factor cubic equations? What methods do you know for dealing with cubics or higher order equations?

If you are allowed to graph this using either a graphing calculator or online (e.g., https://www.desmos.com/calculator), you will be able to see the x-intercepts. That's a good start.