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3 years ago

11

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist

. (enter your answer using interval notation.) t(t − 9)y' + y = 0, y(1) = 1

1 answer:

Olin [163]3 years ago

8 0

If either $t=0$ or $t=9$ , the ODE has the trivial solution $y=0$ , but $y(1)=1\neq0$ . So any solution to the ODE that satisfies the initial condition can only exist on the interval $0$ .

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Could the inverse of a non-function be a function? Explain or give an example.

Kitty [74]

Answer:

The inverse of a non-function mapping is not necessarily a function.

For example, the inverse of the non-function mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ is the same as itself (and thus isn't a function, either.)

Step-by-step explanation:

A mapping is a set of pairs of the form $(a,\, b)$ . The first entry of each pair is the value of the input. The second entry of the pair would be the value of the output.

A mapping is a function if and only if for each possible input value $x$ , at most one of the distinct pairs includes $x\!$ as the value of first entry.

For example, the mapping $\lbrace (0,\, 0),\, (1,\, 0) \rbrace$ is a function. However, the mapping $\lbrace (0,\, 0),\, (1,\, 0),\, (1,\, 1) \rbrace$ isn't a function since more than one of the distinct pairs in this mapping include $1$ as the value of the first entry.

The inverse of a mapping is obtained by interchanging the two entries of each of the pairs. For example, the inverse of the mapping $\lbrace (a_{1},\, b_{1}),\, (a_{2},\, b_{2})\rbrace$ is the mapping $\lbrace (b_{1},\, a_{1}),\, (b_{2},\, a_{2})\rbrace$ .

Consider mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ . This mapping isn't a function since the input value $0$ is the first entry of more than one of the pairs.

Invert $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ as follows:

$(0,\, 0)$ becomes $(0,\, 0)$ .
$(0,\, 1)$ becomes $(1,\, 0)$ .
$(1,\, 0)$ becomes $(0,\, 1)$ .
$(1,\, 1)$ becomes $(1,\, 1)$ .

In other words, the inverse of the mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ would be $\lbrace (0,\, 0),\, (1,\, 0),\, (0,\, 1),\, (1,\, 1) \rbrace\!$ , which is the same as the original mapping. (Mappings are sets. There is no order between entries within a mapping.)

Thus, $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ is an example of a non-function mapping that is still not a function.

More generally, the inverse of non-trivial ellipses (a class of continuous non-function $\mathbb{R} \to \mathbb{R}$ mappings, including circles) are also non-function mappings.

3 0

2 years ago

State the domain restriction(s) in interval notation of \displaystyle f\left(g\left(x\right)\right)f(g(x)) given: \displaystyle

DENIUS [597]

Answer:

The interval notation for the domain is $[\frac{23}{3},\infty ]$ .

Step-by-step explanation:

Consider the provided information.

It is given that $\:f\left(x\right)=\sqrt{3x-2},\:\text{ and }\:g\left(x\right)=x-7$

We need to find the value of $f\left(g\left(x\right)\right)$ .

Put the value of g(x) in $f\left(g\left(x\right)\right)$ .

$f\left(g\left(x\right)\right)=f(x-7)$ ....(1)

Now, put x=x-7 in $\:f\left(x\right)=\sqrt{3x-2}$

$\:f\left(x-7\right)= \sqrt{3(x-7)-2}$

$\:f\left(x-7\right)= \sqrt{3x-21-2}$

$\:f\left(x-7\right)= \sqrt{3x-23}$

From equation 1.

$f\left(g\left(x\right)\right)=\:f\left(x-7\right)= \sqrt{3x-23}$

The domain of the function is the set of input values for which a function is defined.

Here, the value of $3x-23$ should be greater or equal to 0 as the square root of a negative number is not real.

Domain= $3x-23\geq0$

$x\geq\frac{23}{3}$

The value of x is all real number greater than $\frac{23}{3}$ .

Hence, the interval notation for the domain is $[\frac{23}{3},\infty ]$ .

7 0

3 years ago

Given: F(x) = 2x - 1; G(x) = 3x + 2; H(x) = x 2 Find F[G(x)] - F(x). 4x + 4 4x + 2 4x

Allushta [10]

Answer:

4x + 4

Step-by-step explanation:

F(x) = 2x - 1

G(x) = 3x + 2

H(x) = x²

We have to calculate the expression F(G(x)) - F(x)

F(G(x)) means the composition of functions F(x) and G(x). In order to find F(G(x)) we have to replace ever occurrence of x in F(x) with the value of G(x). So,

F(G(x)) = 2(3x+2) - 1 = 6x + 4 - 1 = 6x + 3

Thus,

F(G(x)) - F(x) = 6x + 3 - (2x - 1)

= 6x + 3 -2x + 1

= 4x + 4

Therefore, the expression F(G(x)) - F(x) equals 4x + 4

4 0

3 years ago

Solve the math problem

Colt1911 [192]

Answer:I think its ether y-interspersed or equation

Step-by-step explanation:

6 0

2 years ago

Two triangles have altitudes of equal length. if the areas of these triangles have a ratio of 3:4 then the bases of the triangle

MatroZZZ [7]

Answer:
3/4
Area of the small triangle/Area of larger triangle=3/4

Area of the small triangle/Area of larger triangle=(1/2×base of small triangle×length)/(1/2×base of larger triangle×height)
3/4=(1/2×base of small triangle×height)/(1/2×base of larger triangle×height)
Since the length of the triangle are equal the base ratio of the two triangles is the same as area ratio.

5 0

3 years ago