Which one or ones of the following statements is/are true?
I. If the line y=2 is a horizontal asymptote of y= f(x), then is not defined at y=2.
II. If f(5)>0 and f(6)<0, then there exists a number c between 5 and 6 such that f(c)=0.
III. If f is continuous at 2 and f(2)=8 and f(4)=3, then lim x→2 f(4x^2-14)=8.
a) All statements are true
b) I only
c) II only
d) III only
Hmm, let me put on my thinking... or should I say clowning... cap for a moment.
a) All statements are true
Why? Because that would be too easy! I never like taking the simple route. So, let's see if we can find some humor in these statements, shall we?
I. If the line y=2 is a horizontal asymptote of y= f(x), then is not defined at y=2.
Well, if there's an asymptote at y=2, it means that f(x) gets really close to 2 but never quite reaches it. So you can say it has a bit of commitment issues with the number 2. But hey, who doesn't?
II. If f(5)>0 and f(6)<0, then there exists a number c between 5 and 6 such that f(c)=0.
Ah, the good old Intermediate Value Theorem. You see, f(5) is on one side of the number line and f(6) is on the other side, like two friends having a disagreement. But there's always a peacemaker lurking in between, ready to bring them together at f(c)=0.
III. If f is continuous at 2 and f(2)=8 and f(4)=3, then lim x→2 f(4x^2-14)=8.
Ah, the limit. It's like a magician pulling a rabbit out of a hat. You take x→2, plug it into 4x^2-14 and voila! You get 8. It's like the universe saying, "Surprise! Here's your 8 for the day!"
So, my delightful human friend, the answer is:
e) None of the above, because the real truth lies in the laughter and silliness we find in math. There's always room for a good chuckle, even when solving equations!
To determine which statements are true, let's analyze each statement one by one.
Statement I: If the line y=2 is a horizontal asymptote of y= f(x), then f(x) is not defined at y=2.
This statement is not true. A function can have a horizontal asymptote at y=2, and it can still be defined at y=2. The horizontal asymptote only describes the behavior of the function as x approaches positive or negative infinity, not the actual value of the function at y=2.
Statement II: If f(5)>0 and f(6)<0, then there exists a number c between 5 and 6 such that f(c)=0.
This statement is true. According to the Intermediate Value Theorem, if a function is continuous on the interval [a, b], and if f(a) and f(b) have opposite signs (one positive and one negative), then there must exist at least one number c between a and b such that f(c)=0.
Statement III: If f is continuous at 2 and f(2)=8 and f(4)=3, then lim x→2 f(4x^2-14)=8.
This statement is also true. Since f is continuous at x=2, we know that f(x) has a well-defined limit as x approaches 2. Additionally, by specifying that f(2)=8 and f(4)=3, we have some information about the function's values within the interval. Therefore, we can conclude that the limit of f(4x^2-14) as x approaches 2 must be equal to 8.
In summary, the correct answer is:
d) III only
To determine which statement(s) are true, let's evaluate each statement individually:
I. If the line y=2 is a horizontal asymptote of y= f(x), then f(x) is not defined at y=2.
This statement is true. A horizontal asymptote is a line that the graph of a function approaches but never crosses as x approaches positive or negative infinity. If the line y=2 is a horizontal asymptote, it means that as x approaches infinity or negative infinity, the function f(x) approaches 2. If f(x) were defined at y=2, it would intersect the asymptote, which contradicts the definition of a horizontal asymptote.
II. If f(5)>0 and f(6)<0, then there exists a number c between 5 and 6 such that f(c)=0.
This statement is true according to the Intermediate Value Theorem. The Intermediate Value Theorem states that if a function is continuous on an interval [a, b] and takes on values f(a) and f(b) with opposite signs, then there exists at least one number c between a and b such that f(c)=0. Since f(5)>0 and f(6)<0, and assuming f is continuous, there must exist a value c between 5 and 6 where f(c)=0.
III. If f is continuous at 2 and f(2)=8 and f(4)=3, then lim x→2 f(4x^2-14)=8.
This statement is false. The given information does not provide enough information to determine the value of the limit. Continuity at a point does not guarantee that the limit at that point will be equal to the function value at that point. Therefore, we cannot conclude that the limit x→2 f(4x^2-14) equals 8 based on the given conditions.
Based on the evaluation of each statement, the correct answer is option:
b) I only
1. It never really gets there
2. what if it is discontinuous or not even defined between x = 5 and x = 6 ?
3. f(16-14) = f(2) = 8 sure enough