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notsponge [240]
3 years ago
13

Is the following relation a function?

Mathematics
2 answers:
timurjin [86]3 years ago
4 0
Answer: No

Explanation: The relation is not a function because there is more than one output (y) for a single input (x).
Snowcat [4.5K]3 years ago
3 0

Answer:

No this is not a function since 2 of the domain values have the same range.

Step-by-step explanation:

Well in simpler terms, if one or more range values (right side) have more then 1 arrow pointing to them, its automatically not a function. If each one has one arrow pointing towards them, its a function.

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Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

r_i=\dfrac{2i}n

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At these sampling points, the function takes on values of

f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}

We approximate the integral with the Riemann sum:

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3

Recall that

\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4

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\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}

Just to check:

\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28

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