Suppose f and g are continuous functions such that

g(3) = 4
and
lim
x β†’ 3
[3f(x) + f(x)g(x)] = 35.
Find
f(3).

Well, it seems like f and g are in quite a mathematical relationship here. I guess you could say they're in a "function-ship". 🀑

But let's get serious for a moment. We're given that g(3) = 4, but we want to find f(3). Since f and g are continuous functions, we can use the limit to find f(3).

We're given that:
lim [x β†’ 3] [(3f(x)) + (f(x)g(x))] = 35

Let's break this down a bit. We can rewrite it as:
lim [x β†’ 3] (3f(x)) + lim [x β†’ 3] (f(x)g(x)) = 35

Since g(3) = 4, we can rewrite the second part as:
lim [x β†’ 3] (f(x) * 4)

Now, let's isolate f(x) in the first part:
lim [x β†’ 3] (3f(x)) = 35 - lim [x β†’ 3] (f(x) * 4)

If we simplify the left side, we get:
3 * lim [x β†’ 3] f(x) = 35 - 4 * lim [x β†’ 3] f(x)

Let's call lim [x β†’ 3] f(x) as L. Now our equation becomes:
3L = 35 - 4L

Simplifying further:
7L = 35

Finally, we get:
L = 35/7
L = 5

And there you have it! We found that lim [x β†’ 3] f(x) = 5, which means f(3) = 5. So, f(3) = 5 and we've successfully solved the problem. Hope my "function-ship" didn't confuse you too much! 🀑

To find f(3), we need to make use of the given information and apply the Limit Laws.

From the given limit, we can rewrite it as:
lim(x β†’ 3) [3f(x) + f(x)g(x)] = lim(x β†’ 3) 3f(x) + lim(x β†’ 3) f(x)g(x) = 35.

Since we know g(3) = 4, and g(x) is continuous, we can replace g(x) with its value at x = 3:
lim(x β†’ 3) 3f(x) + lim(x β†’ 3) f(x)g(x) = 3f(3) + f(3)g(3) = 35.

Now, we substitute g(3) = 4 into the equation:
3f(3) + f(3)g(3) = 3f(3) + f(3)(4) = 3f(3) + 4f(3) = 7f(3) = 35.

Simplifying the equation, we have:
7f(3) = 35.

To solve for f(3), we divide both sides of the equation by 7:
f(3) = 35/7 = 5.

Therefore, f(3) = 5.

To find f(3), we can rewrite the limit expression using the given information about g(3) and the limit:

lim
x β†’ 3
[3f(x) + f(x)g(x)] = 35.

Since g(3) = 4, we can substitute 4 for g(x) in the expression:

lim
x β†’ 3
[3f(x) + f(x) * 4] = 35.

Simplifying further:

lim
x β†’ 3
[3f(x) + 4f(x)] = 35.

Combine like terms:

lim
x β†’ 3
[7f(x)] = 35.

Now, we can evaluate the limit:

7f(3) = 35.

Divide both sides of the equation by 7 to solve for f(3):

f(3) = 35/7.

Simplifying:

f(3) = 5.

Therefore, f(3) is equal to 5.

It seems we can assume that

lim(x→3) f(x) = f(3)
In that case, we have

3f(3) + f(3)*4 = 35
7f(3) = 35
f(3) = 5