Answer to Question #242820 in Calculus for anand

Question #242820
please do this question as it is very urgwnt for me and for my assignments.



Trace the curve y = 8/(4-x^2) and state all the properties you use to trace it.
1
Expert's answer
2021-10-18T17:08:59-0400

Solution;


All the properties used to trace the graph are;

1)Domain.

Find the domain of the function and determine points of discontinuity.

f(x)=84x2f(x)=\frac{8}{4-x^2}

Rewrite as follows;

f(x)=8(x24)=8(x+2)(x2)f(x)=\frac{8}{-(x^2-4)}=\frac{8}{-(x+2)(x-2)}

For domain gaps,take the denominator to be equal to zero;

-(x+2)(x-2)=0

Therefore,the points of discontinuity are;

x=2or2x=2 or -2

2) Intercepts.

For y -intercept,take x=0

f(0)=840=2f(0)=\frac{8}{4-0}=2

y-intercept is (0,2)

The curve has no x-intercept.

3) Symmetry of the function.

The curve has neither axially symmetry nor point symmetry.

4)Intervals of increase or decrease.

5)Local maximum and minimum.

For (4) and (5),

f(x)=84x2f(x)=\frac{8}{4-x^2}

The first derivative of f(x);

f(x)=16x(1)2(x+2)2(x2)2f'(x)=\frac{16x}{(-1)^2(x+2)^2(x-2)^2}

The second derivative is;

f(x)=48x4128x2256(x+2)4(x2)4f''(x)=\frac{48x^4-128x^2-256}{-(x+2)^4(x-2)^4}

Looking for turning points;

Equate denominator of f'(x) to 0;

(1)2(x+2)2(x2)2=0(-1)^2(x+2)^2(x-2)^2=0

Hence the gaps are at;

x=2 or -2

If f'(x)=0;

16x(1)2(x+2)2(x2)2=0\frac{16x}{(-1)^2(x+2)^2(x-2)^2}=0

Multiply both sides with the denominator;

16x=0

x=0

Turning points could be at x=0.

Substitute 0 into f''(x);

f(0)=48(0)128(0)256(0+2)4(02)4f''(0)=\frac{48(0)-128(0)-256}{-(0+2)^4(0-2)^4} =1

Check if f'(x) changes sign to confirm;

Use -1 and 1;

f(1)=16(1)2(1+2)2(12)2=1.78f'(1)=\frac{16}{(-1)^2(1+2)^2(1-2)^2}=1.78

f(1)=16×1(1)2(1+2)2(12)2=1.78f'(-1)=\frac{16×-1}{(-1)^2(-1+2)^2(-1-2)^2}=-1.78

There is change from positive to negative,hence there exist a turning point;

Substitute 0 into the f(x);

f(0)=840=0f(0)=\frac{8}{4-0}=0

There is a minimum turning point at ;

(0,2)


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