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Locate the points of discontinuity in the piecewise function shown below.

algol13

Answer:

Step-by-step explanation:

The given piecewise function i

From the given function it is clear that function is divided at x=-1 and x=2. It means we check the discontinuity at x=-1 and x=2.

For x=-1,

LHL:

Since LHL ≠ f(-1), therefore the given function is discontinuous at x=-1.

For x=2,

LHL:

Since LHL ≠ f(2), therefore the given function is discontinuous at x=2.

Therefore, the correct option is A.

8 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

Adam is building a computer desk with a separate compartment for the computer. The compartment for the computer is a rectangular

Alik [6]

Answer:

The width of the computer compartment is 9 inches

The length of the computer compartment is 33 inches

The height of the computer compartment is 27 inches

Step-by-step explanation:

The given data on the rectangular prism compartment are;

The volume of the rectangular prism, V = 8019 cubic inches

The length of the compartment, l = 24 inches + The width, w

The height of the compartment, h = 18 inches + The width, w

Therefore, we have;

l = 24 + w

h = 18 + w

V = l × h × w

∴ V = (24 + w) × (18 + w) × w = w³ + 42·w² + 432·w = 8019

w³ + 42·w² + 432·w - 8019 = 0

By graphing the above function with MS Excel, we have one of the solution is w = 9

∴ (w - 9) is a factor of w³ + 42·w² + 432·w - 8019 = 0

Dividing, we get;

w² + 51·w + 891

w³ + 42·w² + 432·w - 8019 = 0

w³ - 9·w²

51·w² - 459·w

891·w

891·w + 8019

0

Therefore,

w³ + 42·w² + 432·w - 8019 = 0

w³ + 42·w² + 432·w - 8019 = (x - 9)·(w² + 51·w + 891) = 0

The determinant of the factor, w² + 51·w + 891 = 51² - 4×1×891 = -963, therefore, the it has complex roots

Therefore, the real solution of (x - 9)·(w² + 51·w + 891) = 0 is w = 9

The width of the computer compartment, w = 9 inches

The length of the computer compartment, l = 24 inches + 9 inches = 33 inches

The height of the computer compartment, h = 18 inches + 9 inches = 27 inches.

3 0

3 years ago

How do u do long division

Archy [21]

can you send me the actual problem so that i can solve it

8 0

3 years ago

The population of a parish is growing with an annual percentage rate compounded continuously. The population reaches 1.1 times i

coldgirl [10]

Answer:

Growth rate = ln(1.1)/2

Step-by-step explanation:

Given:

P1 = 1.1 P0

Number of year n = 2

Computation:

P1 = P0[eⁿˣ]

x = growth rate

So,

1.1 P0 = P0[eⁿˣ]

1.1 = eⁿˣ

1.1 = e²ˣ

By taking log

Growth rate = ln(1.1)/2

6 0

3 years ago