Let f(x) = {(x³+x²-16x+20)/(x-2)² if x ≠ 2, k if x = 2}. If f(x) is continuous for all x, then k = .

Q: Wait, why do we cancel $(x-2)^2$ if $x=2$ makes the denominator zero?

That is a fantastic question! Remember the definition of a limit: we are looking at the behavior of the function as $x$ gets *arbitrarily close* to $2$, but $x$ is *never actually equal* to $2$. Since $x \neq 2$, the term $(x-2)^2$ is never zero, which allows us to safely cancel it out.

Q: Could I have used L'Hopital's Rule instead of factoring?

Absolutely! Since you have a $\frac{0}{0}$ indeterminate form, you can differentiate the numerator and the denominator with respect to $x$. However, because the denominator is $(x-2)^2$, you would need to apply the rule twice. Factoring is often more elegant and less prone to calculation errors in these types of problems.

Q: What if the limit didn't exist? Would the function still be continuous?

No. For a function to be continuous at a point, the limit *must* exist. If the left-hand limit and right-hand limit were different, or if the limit approached infinity, the function would have a jump or an infinite discontinuity, and no value of $k$ could 'fix' it.

Question

Let $f(x) = \begin{cases} \frac{(x^3 + x^2 - 16x + 20)}{(x-2)^2} & 	ext{if } x 
eq 2 \ k & 	ext{if } x = 2 \end{cases}$. If $f(x)$ is continuous for all $x$, then $k = .........$

Sigma Percentile · Accepted Answer

### The Quest for Continuity: Filling the Pothole

Imagine you are driving down a perfectly smooth, winding road. This road represents a continuous function—a path where you can trace the entire curve without ever lifting your pen from the paper.

But suddenly, you encounter a pothole. In the world of calculus, that pothole is a point of discontinuity. Our goal today is to find the exact amount of 'asphalt'—the value $k$—needed to fill that hole and make our road perfectly smooth again.

### The Mathematical Landscape

We are presented with a piecewise function:

$$f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x-2)^2} & 	ext{if } x 
eq 2 \ k & 	ext{if } x = 2 \end{cases}$$

Our mission is to ensure that the function is continuous for all $x$. The definition of continuity at a point $x=a$ is elegant and precise: the limit of the function as $x$ approaches $a$ must equal the value of the function at $a$.

In our case, we need:

$$\lim_{x 	o 2} f(x) = f(2) = k$$

This is our North Star. If we can find the value of that limit, we have found our $k$.

### The Indeterminate Trap

Let’s try the most direct approach: substitution. If we plug $x=2$ into our rational expression, we get:

$$\frac{2^3 + 2^2 - 16(2) + 20}{(2-2)^2} = \frac{8 + 4 - 32 + 20}{0} = \frac{0}{0}$$

We have arrived at the classic $\frac{0}{0}$ indeterminate form. Do not panic! This is not a dead end; it is a signpost.

It tells us that the factor $(x-2)$ is hiding in the numerator, waiting to be revealed. The denominator is already $(x-2)^2$, so we know that if we can factor $(x-2)^2$ out of the numerator, the mystery will be solved.

### The Algebraic Excavation

We need to factor the cubic polynomial $x^3 + x^2 - 16x + 20$. Since we know $(x-2)$ is a factor, we can perform polynomial division.

Dividing $x^3 + x^2 - 16x + 20$ by $(x-2)$ yields the quadratic $x^2 + 3x - 10$. Now, we factor that quadratic:

$$x^2 + 3x - 10 = (x-2)(x+5)$$

Putting it all together, the numerator is:

$$x^3 + x^2 - 16x + 20 = (x-2)(x-2)(x+5) = (x-2)^2(x+5)$$

### The Grand Cancellation

Now, look at the beauty of the expression when we substitute this back into our limit:

$$k = \lim_{x 	o 2} \frac{(x-2)^2(x+5)}{(x-2)^2}$$

Because we are taking the limit as $x$ approaches $2$ (and not *at* $2$), we know that $x 
eq 2$. This means $(x-2)^2$ is non-zero, allowing us to cancel it from the numerator and denominator.

The expression simplifies dramatically to:

$$k = \lim_{x 	o 2} (x+5)$$

### The Final Stretch

With the problematic terms gone, the limit becomes trivial. We simply evaluate the expression at $x=2$:

$$k = 2 + 5 = 7$$

And there it is! By setting **$k=7$**, we have effectively filled the pothole. The function is now continuous at $x=2$, and the road is smooth once more.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = {\frac{( x ^{3} + x ^{2} - 16 x + 20 )}{( x - 2 ) ^{2}} k if x \neq = 2 if x = 2$ . If $f (x)$ is continuous for all $x$ , then $k = .........$

Enter Numerical Value:

Visualized Solution

Continuity at $x = 2$

The Limit Condition

Setting up the Limit

Identifying the $\frac{0}{0}$ Form

Factorizing the Numerator (Part 1)

Factorizing the Numerator (Part 2)

Fully Factored Numerator

Simplifying the Expression

Canceling Common Terms

Evaluating the Limit

Final Value of $k$

The Sigma Insight: Continuity at a Point and in an Interval

The Quest for Continuity

The Mathematical Landscape

The Indeterminate Trap

The Algebraic Excavation

The Grand Cancellation

The Final Stretch

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