Question

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

Answer 1

The function `f(x)` is continuous at point `a` if and only if the limit
`lim_(x->a) f(x)` exists and equals `f(a)`. So to prove that a function is continuous first you have to calculate the limit

`lim_(x->-1) (x+2x^3)^4=(-1+(-1)^3)^4=(-2)^4=16`

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

`f(-1)=(-1+(-1)^3)^4=(-2)^4=16`

The calculation shows, that `f(-1)=lim_(x->-1)f(x)`, so we can write, that function `f(x)` is continuous at point `-1`

Answer 2

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(x)=(x+2x^3)^4` can be expanded using the Binomial Theorem as:

`f(x)=x^4+ 4x^{3}\cdot 2x^3+6x^2\cdot (2x^3)^2+4x\cdot (2x^3)^3+(2x^3)^4`

`=x^4+8x^6+24x^8+32x^10+16x^12`. Also note that `f(-1)=81`.

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

Without loss of generality, we can assume that `|x+1|=|x-(-1)|<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|\leq |x+1|+|-1|<1+1=2` so that: 1) `|4x^6+4x^4+x^2+9|\leq 4|x|^6+4|x|^4+|x|^2+9<4\cdot 2^6+4\cdot 2^4+2^2+9=333`, 2) `|2x^2+2x+3|\leq 2|x|^2+2|x|+|3|<2\cdot 2^2+2\cdot 2+3=15`, 3) `|2x^2-2x+3|\leq 2|x|^2+2|x|+|3|<2\cdot 2^2+2\cdot 2+3=15`, and 4) `|x-1|\leq |x|+|1|<2+1=3`. Putting these all together means `|f(x)-81|<333\cdot 15\cdot 15\cdot 3\cdot |x+1|=224775\cdot |x+1|` when `|x+1|<1`.

Somebody please let me know if I made any mistakes.

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

From this point, do the same kind of work as shown above (fill in the gaps `\cdots`) to say that `|f(x)-81|=\cdots<224775\cdot |x+1|<224775\cdot \delta\leq 224775\cdot 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).

Answer 3

"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_(xrarr-1) f(x) = f(-1)`

First attempt
Have you proven that all polynomials are continuous on their domains ?
If so, then point out that `f(x) = (x+2x^3)^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_(xrarr-1)(x+2x^3)^4 = [lim_(xrarr-1)(x+2x^3)]^4` `color(white)"sss"` (power)

`= [lim_(xrarr-1)(x)+ lim_(xrarr-1)(2x^3)]^4` `color(white)"sss"` (sum)

`= [lim_(xrarr-1)(x)+ 2lim_(xrarr-1)(x^3)]^4` `color(white)"sss"` (constant multiple)

`= [lim_(xrarr-1)(x)+ 2(lim_(xrarr-1)(x))^3]^4` `color(white)"sss"` (power)

`= [(-1)+2(-1)^3]^4` `color(white)"ssssss"` (`lim_(xrarra) x = a`)

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

`f(-1) = [(-1)+2(-1)^3]^4 = 81`.

That is: `lim_(xrarr-1)f(x) = f(-1)`

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.