Calculate the following:
a) Find g(f(1))
b) Find lim x-> 1+ g(f(x))
c) Find f(g(0))
d) Find lim x->0- f(g(x))
f(x)= 1-x, x<1
1, x=1
x-1, x>1
g(x)= -x, x<0
2, x=0
x+2, x>0
This is where it gets hard. I have a graph for f(x) and one for g(x). I will explain them. The graph for f(x)has an open circle at 1 on the x-axis and a line extending straight out from it going to the right. Another line extends out from it as well in the left direction and passes through (0,1). A closed circle is located at (1,1).
The graph for g(x) has an open circle at the origin with a line extending from it going left. A closed circle is located at (0,2) with a line extending from it going to the right.
I have been struggling with these and I am now also confused as to why he included the info with the x is <,=,> part, as I have never seen this before. Could someone please help. I would really appreciate it.
a) To find g(f(1)), we need to substitute 1 into the function f(x) first, then take the result and substitute it into the function g(x).
First, let's find f(1):
f(1) = 1 - 1 = 0
Now, substitute f(1) into g(x):
g(f(1)) = g(0)
Now, we need to evaluate g(x) for x = 0:
g(0) = 2
Therefore, g(f(1)) = 2.
b) To find lim x->1+ g(f(x)), we need to evaluate the limit of g(f(x)) as x approaches 1 from the right.
As x approaches 1 from the right, the value of f(x) approaches 1-1 = 0.
Now, substitute f(x) = 0 into g(x):
g(f(x)) = g(0)
As we found earlier, g(0) = 2. Therefore, lim x->1+ g(f(x)) = 2.
c) To find f(g(0)), we need to substitute 0 into the function g(x) first, then take the result and substitute it into the function f(x).
First, let's find g(0):
g(0) = 2
Now, substitute g(0) into f(x):
f(g(0)) = f(2)
Now, we need to evaluate f(x) for x = 2:
f(2) = 2 - 1 = 1
Therefore, f(g(0)) = 1.
d) To find lim x->0- f(g(x)), we need to evaluate the limit of f(g(x)) as x approaches 0 from the left.
As x approaches 0 from the left, the value of g(x) approaches -x.
Now, substitute g(x) = -x into f(x):
f(g(x)) = f(-x)
Now, we need to evaluate f(x) for x = -x:
f(-x) = 1 - (-x) = 1 + x
Therefore, lim x->0- f(g(x)) = lim x->0- (1 + x) = 1.
I hope my attempts at humor have made navigating through these calculations a bit more enjoyable for you! Let me know if there's anything else I can assist you with.
a) To find g(f(1)), we need to evaluate f(1) first, and then substitute this value into g(x).
Evaluate f(1):
Since 1 = 1, f(1) = 1.
Substitute f(1) into g(x):
g(f(1)) = g(1).
Next, we need to evaluate g(1). From the information given, g(x) = x+2 for x > 0. Since 1 > 0, we can substitute 1 into g(x):
g(1) = 1 + 2 = 3
Therefore, g(f(1)) = g(1) = 3.
b) To find lim x->1+ g(f(x)), we need to evaluate the limit of g(f(x)) as x approaches 1 from the right.
First, let's calculate f(x) for values of x approaching 1 from the right:
lim x->1+ f(x) = lim x->1+ (x-1) = 1-1 = 0
Now, we substitute this value into g(x):
lim x->1+ g(f(x)) = lim x->1+ g(0)
From the information given, g(x) = 2 for x = 0. Therefore:
lim x->1+ g(f(x)) = lim x->1+ g(0) = g(0) = 2
c) To find f(g(0)), we need to evaluate g(0) first, and then substitute this value into f(x).
From the information given, g(x) = 2 for x = 0:
g(0) = 2
Now, we substitute g(0) into f(x):
f(g(0)) = f(2).
From the information given, f(x) = x-1 for x > 1. Since 2 is greater than 1, we can substitute 2 into f(x):
f(g(0)) = f(2) = 2-1 = 1
d) To find lim x->0- f(g(x)), we need to evaluate the limit of f(g(x)) as x approaches 0 from the left.
First, let's calculate g(x) for values of x approaching 0 from the left:
lim x->0- g(x) = lim x->0- (-x) = 0 (since -0 = 0)
Now, we substitute this value into f(x):
lim x->0- f(g(x)) = lim x->0- f(0).
From the information given, f(x) = 1-x for x < 1. Since 0 is less than 1, we can substitute 0 into f(x):
lim x->0- f(g(x)) = lim x->0- f(0) = f(0) = 1-0 = 1
Therefore, lim x->0- f(g(x)) = 1.
Sure, I'd be happy to help you with these calculations. The functions f(x) and g(x) are defined piecewise, which means that they have different rules depending on the value of x. In this case, f(x) has different rules for x<1, x=1, and x>1, and g(x) has different rules for x<0, x=0, and x>0.
Let's go through the calculations step by step:
a) To find g(f(1)), we need to substitute the value 1 into the function f(x) and then substitute the resulting value into the function g(x).
We start by evaluating f(1):
f(1) = 1 - 1 = 0
Now we substitute the value f(1) = 0 into g(x):
g(f(1)) = g(0)
Looking at the definition of g(x), we see that g(0) is equal to 2. Therefore, g(f(1)) = g(0) = 2.
b) To find the limit lim x-> 1+ g(f(x)), we need to evaluate g(f(x)) as x approaches 1 from the right.
As x approaches 1 from the right, f(x) approaches 1 since f(x) = 1 when x = 1. Now let's substitute this value into g(x):
lim x-> 1+ g(f(x)) = lim x-> 1+ g(1)
From the definition of g(x), we can see that g(1) is equal to 2. Therefore, lim x-> 1+ g(f(x)) = lim x-> 1+ g(1) = 2.
c) To find f(g(0)), we need to substitute the value 0 into the function g(x) and then substitute the resulting value into the function f(x).
Evaluating g(0):
g(0) = 2
Now we substitute the value g(0) = 2 into f(x):
f(g(0)) = f(2)
Looking at the definition of f(x), we can see that f(x) is equal to x-1 when x > 1. Since g(0) = 2 > 1, the value of f(x) when x = 2 is:
f(2) = 2 - 1 = 1
Therefore, f(g(0)) = f(2) = 1.
d) To find the limit lim x->0- f(g(x)), we need to evaluate f(g(x)) as x approaches 0 from the left.
As x approaches 0 from the left, g(x) approaches 0 since g(x) = 0 when x = 0. Now let's substitute this value into f(x):
lim x->0- f(g(x)) = lim x->0- f(0)
From the definition of f(x), we can see that f(x) is equal to 1 - x when x < 1. Since g(0) = 0 < 1, the value of f(x) when x = 0 is:
f(0) = 1 - 0 = 1
Therefore, lim x->0- f(g(x)) = lim x->0- f(0) = 1.
I hope this helps! If you have any further questions, please let me know.