I'm finding difficulty in answering this question. Any help would be appreciated.

For the equation,
y1= x^-5x+6 and its reciprocal:

Discuss what is the consequences for y1 to 1/y1 as x --> +- infinity?

Thanks in advance.

clearly as x becomes larger , x^2-5x+6 becomes larger even faster, so dividing by a larger and larger number will make 1/y1 approach zero

who r u TK, r u from NMHS?

Reiny, how could i further prove this?

TP, NMHS? sorry>?

Just out of curiousity, are you from Adelaide? Because I got this same assignment, which is from the SSABSA website. I just thought you were a friend of mine, because my friend also uses the initials TK< but he didn't post this.

LOL @ comments above!!! guessing its Anthony or something ;D

I'm guessing BK is around here somewhere

beyonce Knowles? sweet!..im sensing a mongol...:D and a poon..

I'm thinking of something along the lines of Bor.....K....ien

To understand the consequences of y1 as x approaches positive or negative infinity, we need to examine its behavior as x gets extremely large or small.

To start, let's simplify the equation y1 = x^-5x + 6. The exponent -5x can be rewritten as 1/(5x), so the equation becomes y1 = x^(1/(5x)) + 6.

Now, consider what happens as x approaches positive or negative infinity.

1. As x approaches positive infinity:
When x gets larger and larger, the term 1/(5x) approaches zero. In other words, the exponent approaches zero. This means that x^(1/(5x)) approaches 1. The equation y1 = x^(1/(5x)) + 6 will then approach 1 + 6 = 7 as x approaches positive infinity.

2. As x approaches negative infinity:
When x gets extremely negative, the term 1/(5x) becomes very large. In this case, x^(1/(5x)) will approach 0. The equation y1 = x^(1/(5x)) + 6 will approach 0 + 6 = 6 as x approaches negative infinity.

Now, let's discuss the consequences for y1 to 1/y1 as x approaches positive or negative infinity.

The reciprocal of y1 is given by 1/y1. So, we need to compute the reciprocal of y1 as x approaches positive and negative infinity.

1. As x approaches positive infinity:
As we mentioned earlier, y1 approaches 7 as x approaches positive infinity. Therefore, 1/y1 will approach 1/7.

2. As x approaches negative infinity:
Similarly, y1 approaches 6 as x approaches negative infinity. Hence, 1/y1 will approach 1/6.

To summarize, as x approaches positive infinity, y1 approaches 7 and 1/y1 approaches 1/7. As x approaches negative infinity, y1 approaches 6 and 1/y1 approaches 1/6.