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4 years ago

Please help me solve this

Mathematics

1 answer:

MakcuM [25]4 years ago

8 0

Answer:

a) Function? No

b) Domain: $-4\le x\le 2$

c) Range: $-3\le y\le 3$

Step-by-step explanation:

a) Function? No

We want to determine whether the graph of the relation is a function or not.

When we draw a straight line at x=-1, it will intersect the graph at several points, therefore the graph fails the vertical line test and hence it is not a function.

b) The domain refers to all x-values for which the function is defined.

From the graph the domain is $-4\le x\le 2$ because the ends of the graph are closed.

c) The range is the set of all y-values for which the graph is defined.

From the graph the range is $-3\le y\le 3$

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Find the area of the region enclosed by the graphs of the functions

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Answer:

$\displaystyle A = \frac{8}{21}$

General Formulas and Concepts:

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Order of Operations: BPEMDAS

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Area - Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

*Note:

Remember that for the Area of a Region, it is top function minus bottom function.

Step 1: Define

f(x) = x²

g(x) = x⁶

Bounded (Partitioned) by x-axis

Step 2: Identify Bounds of Integration

Find where the functions intersect (x-values) to determine the bounds of integration.

Simply graph the functions to see where the functions intersect (See Graph Attachment).

Interval: [-1, 1]

Lower bound: -1

Upper Bound: 1

Step 3: Find Area of Region

Integration

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^1_{-1} {[x^2 - x^6]} \, dx$
[Area] Rewrite [Integration Property - Subtraction]: $\displaystyle A = \int\limits^1_{-1} {x^2} \, dx - \int\limits^1_{-1} {x^6} \, dx$
[Area] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = \frac{x^3}{3} \bigg| \limit^1_{-1} - \frac{x^7}{7} \bigg| \limit^1_{-1}$
[Area] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = \frac{2}{3} - \frac{2}{7}$
[Area] Subtract: $\displaystyle A = \frac{8}{21}$