How do you show the function is continuous at the given number a f(x)=(x+2x^3)^4, a=-1?

May 19, 2015

The function $f \left(x\right)$ is continuous at point $a$ if and only if the limit
${\lim}_{x \to a} f \left(x\right)$ exists and equals $f \left(a\right)$. So to prove that a function is continuous first you have to calculate the limit

${\lim}_{x \to - 1} {\left(x + 2 {x}^{3}\right)}^{4} = {\left(- 1 + {\left(- 1\right)}^{3}\right)}^{4} = {\left(- 2\right)}^{4} = 16$

The limit exists, so now you have to calculate $f \left(- 1\right)$ and check if the value equals the limit

$f \left(- 1\right) = {\left(- 1 + {\left(- 1\right)}^{3}\right)}^{4} = {\left(- 2\right)}^{4} = 16$

The calculation shows, that $f \left(- 1\right) = {\lim}_{x \to - 1} f \left(x\right)$, so we can write, that function $f \left(x\right)$ is continuous at point $- 1$

May 22, 2015

Okay. Maybe I should do an epsilon/delta proof here, though it's not pleasant. It can be done, though you might not want to do it unless your life depended on it (lol).

First, a bit of scratchwork. The function $f \left(x\right) = {\left(x + 2 {x}^{3}\right)}^{4}$ can be expanded using the Binomial Theorem as:

$f \left(x\right) = {x}^{4} + 4 {x}^{3} \setminus \cdot 2 {x}^{3} + 6 {x}^{2} \setminus \cdot {\left(2 {x}^{3}\right)}^{2} + 4 x \setminus \cdot {\left(2 {x}^{3}\right)}^{3} + {\left(2 {x}^{3}\right)}^{4}$

$= {x}^{4} + 8 {x}^{6} + 24 {x}^{8} + 32 {x}^{10} + 16 {x}^{12}$. Also note that $f \left(- 1\right) = 81$.

The distance between $f \left(x\right)$ and $f \left(- 1\right) = 81$ is $| f \left(x\right) - 81 | = | 16 {x}^{12} + 32 {x}^{10} + 24 {x}^{8} + 8 {x}^{6} + {x}^{4} - 81 |$. Through the help of some technology (use "Factor" on Wolfram Alpha), you can find that
$| f \left(x\right) - 81 | = | 4 {x}^{6} + 4 {x}^{4} + {x}^{2} + 9 | \setminus \cdot | 2 {x}^{2} + 2 x + 3 | \setminus \cdot | 2 {x}^{2} - 2 x + 3 | \setminus \cdot | x - 1 | \setminus \cdot | x + 1 |$.

Without loss of generality, we can assume that $| x + 1 | = | x - \left(- 1\right) | < 1$ so that $x$ is within 1 unit of $a = - 1$. Based on this assumption, we can then use the Triangle Inequality (a number of times) to say $| x | = | x + 1 - 1 | \setminus \le q | x + 1 | + | - 1 | < 1 + 1 = 2$ so that: 1) $| 4 {x}^{6} + 4 {x}^{4} + {x}^{2} + 9 | \setminus \le q 4 | x {|}^{6} + 4 | x {|}^{4} + | x {|}^{2} + 9 < 4 \setminus \cdot {2}^{6} + 4 \setminus \cdot {2}^{4} + {2}^{2} + 9 = 333$, 2) $| 2 {x}^{2} + 2 x + 3 | \setminus \le q 2 | x {|}^{2} + 2 | x | + | 3 | < 2 \setminus \cdot {2}^{2} + 2 \setminus \cdot 2 + 3 = 15$, 3) $| 2 {x}^{2} - 2 x + 3 | \setminus \le q 2 | x {|}^{2} + 2 | x | + | 3 | < 2 \setminus \cdot {2}^{2} + 2 \setminus \cdot 2 + 3 = 15$, and 4) $| x - 1 | \setminus \le q | x | + | 1 | < 2 + 1 = 3$. Putting these all together means $| f \left(x\right) - 81 | < 333 \setminus \cdot 15 \setminus \cdot 15 \setminus \cdot 3 \setminus \cdot | x + 1 | = 224775 \setminus \cdot | x + 1 |$ when $| x + 1 | < 1$.

Now here's an outline of the proof. Let $\setminus \epsilon > 0$ be given. Choose $\setminus \delta = \min \left\{1 , \setminus \frac{\epsilon}{224775}\right\} > 0$ (or any positive number smaller than this...haha) and suppose $| x - a | = | x - \left(- 1\right) | = | x + 1 | < \setminus \delta$. Since $| x + 1 | < 1$, it follows by the Triangle Inequality (as above), that $| x | = | x + 1 - 1 | \setminus \le q | x + 1 | + | - 1 | < 1 + 1 = 2$.

From this point, do the same kind of work as shown above (fill in the gaps $\setminus \cdots$) to say that $| f \left(x\right) - 81 | = \setminus \cdots < 224775 \setminus \cdot | x + 1 | < 224775 \setminus \cdot \setminus \delta \setminus \le q 224775 \setminus \cdot \frac{\epsilon}{224775} = \epsilon$.

That finishes the proof. It's nasty problems like this that give abstract theorems their value (like the abstract fact that any polynomial is continuous for all $x$, which can be built up from fairly simple pieces, such as the sum and product of two continuous functions being continuous, as well as the use of mathematical induction).

May 23, 2015

"How you show" depends on what tools you have to work with.

I assmue that you know that $f$ is continuous at $- 1$ if and only if ${\lim}_{x \rightarrow - 1} f \left(x\right) = f \left(- 1\right)$

First attempt
Have you proven that all polynomials are continuous on their domains ?
If so, then point out that $f \left(x\right) = {\left(x + 2 {x}^{3}\right)}^{4}$ is (when expanded) a polynomial (of degree 12). Then point out that $- 1$ is in the domain, so $f$ is continuous at $- 1$.

Second attempt
If you do not have that tool to work with, have you got:
The limit of a power is the power of the limit AND the limit of a constant multiple is the constant multiple times the limit AND the limit of a sum is the sum of the limit. (These often appear as "Limit Laws" or "Properties of Limits")?

If so:

${\lim}_{x \rightarrow - 1} {\left(x + 2 {x}^{3}\right)}^{4} = {\left[{\lim}_{x \rightarrow - 1} \left(x + 2 {x}^{3}\right)\right]}^{4}$ $\textcolor{w h i t e}{\text{sss}}$ (power)

$= {\left[{\lim}_{x \rightarrow - 1} \left(x\right) + {\lim}_{x \rightarrow - 1} \left(2 {x}^{3}\right)\right]}^{4}$ $\textcolor{w h i t e}{\text{sss}}$ (sum)

$= {\left[{\lim}_{x \rightarrow - 1} \left(x\right) + 2 {\lim}_{x \rightarrow - 1} \left({x}^{3}\right)\right]}^{4}$ $\textcolor{w h i t e}{\text{sss}}$ (constant multiple)

$= {\left[{\lim}_{x \rightarrow - 1} \left(x\right) + 2 {\left({\lim}_{x \rightarrow - 1} \left(x\right)\right)}^{3}\right]}^{4}$$\textcolor{w h i t e}{\text{sss}}$ (power)

$= {\left[\left(- 1\right) + 2 {\left(- 1\right)}^{3}\right]}^{4}$ $\textcolor{w h i t e}{\text{ssssss}}$ (${\lim}_{x \rightarrow a} x = a$)

$= 81$
The last line above is clearly equal to:

$f \left(- 1\right) = {\left[\left(- 1\right) + 2 {\left(- 1\right)}^{3}\right]}^{4} = 81$.

That is: ${\lim}_{x \rightarrow - 1} f \left(x\right) = f \left(- 1\right)$

So, by the definition of "continuous at $a$", $f$ is continuous at $a = - 1$.

Third
If you have none of these , you'll need an answer like Bill Kinney's answer.