Calculus Plz Help

Question

If possible, choose so that the following function is continuous on any interval: (piecewise function)

f(x)= (6x^4-18x^3)/(x-3) x(=/)3 ------> the =/ is the equal sign with a slash through it

k x=3

k=

Answer 1

Hi Kyle,

Tricky problem! Let's first just review what the problem means by the function being continuous.

A function f(x) is said to be continuous at x = a if the limit of f(x) as x approaches "a" is equal to f(a). A function is continuous on an interval [a,b] if the function is continuous at all points within the interval.

(Source: http://tutorial.math.lamar.edu/Classes/CalcI/Continuity.aspx) .

OK, so that's a lot to process. So an easier way to test if a function is continuous is to draw it out and then trace along the curve. If at any point you need to lift your pencil when tracing the function, it's not a continuous function. That could be because a function is undefined at a point, or has an asymptote, or there's a jump in the curve, etc.

So now let's look at your function and focus on just the first part:

f(x) = (6x⁴ - 18x³)/(x-3), if x does not equal 3 --> To determine if this is continuous, we need to look for any points where the function may be undefined or jump or somehow break -- somewhere that we need to lift our pencil if we draw it out. Are there any points where this happens?

Yes! We get a clue from the fact that we only use this function when x is not 3. If we try and plug 3 into the function, we end up with 0 on the denominator. If you recall, a number can't be divided by 0 -- it's undefined at that point!

But here's the question: Is that point an asymptote? Is it a hole in the graph? What is it?

Let's draw out the function to find out -- you can do that on your graphing calculator or at this website (https://www.desmos.com/calculator). What do you find?

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You should find that there's a hole in the graph when x = 3.

So the question now is why is it a hole? This is where it gets tricky. It's because the numerator in our function can be simplified:

f(x) = (6x⁴ - 18x³)/(x-3)

= [6x³ (x - 3)]/(x-3) --> [6x³ (x - 3)]/(x-3)

= 6x³

So except for when x = 3, the function is pretty much the same as f(x) = 6x³

Finally we can answer the original problem. Can we find a function or number "k" so that we never have to pick up our pencil when drawing this curve, meaning it's continuous everywhere?

Yes we can! We know that our function is pretty much f(x) = 6x³ except at x = 3 because of that pesky denominator in the unsimplified form of the function. But the simplified form f(x) = 6x³ is continuous everywhere, including at x = 3. So what if we just found what this simplified form of f(x) is when x = 3?

f(3) = 6(3³) = 162

That's our answer! If we let k = 162 when x = 3 for our piecewise function, it fills in the hole of f(x) = (6x⁴ - 18x³)/(x-3). It makes it continuous! If you don't see why, try drawing out the piecewise function for yourself.

Answer 2

Choose k=162

Note the f(x) for x≠3 can be factored so that x-3 divides out leaving 6x³.