Function help! Given f(x) = 3k+2, find k so lim x->6 f(x) = f(6)

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Kaanha

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Okay, I'm helping someone right now do an f(x) problem but it's been YEARS since I did this.

What we have is f(x) = 3k+2

The question is find the value of k such that lim x->6 f(x) = f(6)

Now I've done two things here...first I graphed it, which if I did it right (maybe I didn't) I get a straight line on 2 meaning k = 2?

then I plugged in the values. f(x) = y = 6, So 6=3k+2, solving that I get 0

Could someone help me with this, everything I keep looking online to refresh myself with is WAY more complicated than this problem and I've yet to find anything that just gives me the basics of "solve it like this".

Thank you so much in advance for any help!
 
Last edited:
Kaanha said:
Okay, I'm helping someone right now do an f(x) problem but it's been YEARS since I did this.

What we have is f(x) = 3k+2

The question is find the value of k such that lim x->6 f(x) = f(6)

Now I've done two things here...first I graphed it, which if I did it right (maybe I didn't) I get a straight line on 2 meaning k = 2?

then I plugged in the values. f(x) = y = 6, So 6=3k+2, solving that I get 0

Could someone help me with this, everything I keep looking online to refresh myself with is WAY more complicated than this problem and I've yet to find anything that just gives me the basics of "solve it like this".

Thank you so much in advance for any ehlpe!
Click to expand...
As I understand your question:

Given f(x) = 3k+2 find the value of k such that lim x->6 f(x) = f(6).

However, (3k +2) is not a function of 'x'.

Please check your original problem and post for correctness!



 
That...is the original problem.

Here is a picture fo the actual assignment, it's B) we're working on.

IMG_20181010_074248679.jpg
 
That was the correct problem.

Here is an image of the actual homework page, we're working on B)

IMG_20181010_074248679.jpg
 
I apologize to any moderator if I did this wrong, I've tried posting this twice now and it doesn't seem to be going up, I'm not sure if it's because of the image or not so I resized it.

I copied the question exactly so it's correct. I've included a snippet from the actual homework page here so you can see it directly, it's Question 2b.

Homework Snippet.jpg
 
Kaanha said:
Okay, I'm helping someone right now do an f(x) problem but it's been YEARS since I did this.

What we have is f(x) = 3k+2

The question is find the value of k such that lim x->6 f(x) = f(6)

Now I've done two things here...first I graphed it, which if I did it right (maybe I didn't) I get a straight line on 2 meaning k = 2?

then I plugged in the values. f(x) = y = 6, So 6=3k+2, solving that I get 0

Could someone help me with this, everything I keep looking online to refresh myself with is WAY more complicated than this problem and I've yet to find anything that just gives me the basics of "solve it like this".

Thank you so much in advance for any help!
Click to expand...
Since f(x) = 3k+2 is a constant function of x, then lim x->6 f(x) = f(6) = 3k+2. This is correct for any value of k.
 
Find limit as x approaches to 6 first

Step 1:

x2 - 5x - 6 (x-6) (x+1)
lim (x=>6) ____________ = lim (x=>6) ____________ = lim (x=>6) (x+1) = 7
x-6 (x-6)



Around x = 6, the function (x2 - 5x - 6) approaches to y = 7.


Step 2:

So when x = 6, it must be y = 7

Therefore;

3k + 2 = 7
k = 5
3




Kaanha said:
Okay, I'm helping someone right now do an f(x) problem but it's been YEARS since I did this.

What we have is f(x) = 3k+2

The question is find the value of k such that lim x->6 f(x) = f(6)

Now I've done two things here...first I graphed it, which if I did it right (maybe I didn't) I get a straight line on 2 meaning k = 2?

then I plugged in the values. f(x) = y = 6, So 6=3k+2, solving that I get 0

Could someone help me with this, everything I keep looking online to refresh myself with is WAY more complicated than this problem and I've yet to find anything that just gives me the basics of "solve it like this".

Thank you so much in advance for any help!
Click to expand...
 
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