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Evgesh-ka [11]
3 years ago
15

For what value of c is the function defined below continuous on (-\infty,\infty)?

Mathematics
1 answer:
kozerog [31]3 years ago
4 0
f(x)= \left \{ {{x^2-c^2,x \ \textless \  4} \atop {cx+20},x \geq 4} \right


It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if 
</span><span>\lim_{x \rightarrow 4} \  f(x) = f(4)

</span><span>In notation we write respectively
</span>\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence 
\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2

Thus these two limits, the one from above and below are equal if and only if
 4c + 20 = 16 - c²<span> 
 Or in other words, the limit as x --> 4 of f(x) exists if and only if
 4c + 20 = 16 - c</span>²

c^2+4c+4=0&#10;\\(c+2)^2=0&#10;\\c=-2

That is to say, if c = -2, f(x) is continuous at x = 4. 

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers (-\infty, +\infty)

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25 POINTS AVAILABLE
myrzilka [38]

Answer:

\large\boxed{1.\ (-3, 0),\ r = 3}\\\boxed{2.\ (x+4)^2+(y-3)^2=36}\\\boxed{3.\ (x-2)^2+(y-1)^2=(\sqrt{34})^2}

Step-by-step explanation:

The equation of a circle in standard form:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

1. We have the equation:

(x+3)^2+y^2=9\\\\(x-(-3))^2+(y-0)^2=3^2

<h2 />

2. We have the center (-4, 3) and the radius r = 6. Substitute:

(x-(-4))^2+(y-3)^2=6^2\\\\(x+4)^2+(y-3)^2=36

3. We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)

Substitute:

h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substitute the coordinates of the points (2, 1) and (5, -4):

r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}

Finally we have:

(x-2)^2+(y-1)^2=(\sqrt{34})^2

7 0
3 years ago
What is the difference between 7/10-44/100
katrin2010 [14]

Answer:

7/10-44/100                                                                                                                            70-44

 100

=26/100

=13/50

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
48+49+50+51+...112+113
Stels [109]

Answer: 423

Step-by-step explanation:

4 0
3 years ago
Write the sum of 18+27 as the product of their gcf and another sum
Serjik [45]

One <em>possible answer</em> is:

9×(2+3)

Explanation:

First we find the sum of 18 and 27. 18+27 = 45.

Next we find the GCF of 18 and 27. We do this by finding the prime factorization of each:

18 = 2*9

9 = 3*3

Thus, 18 = 2*3*3.

27 = 3*9

9 = 3*3

Thus, 27 = 3*3*3

We want the common factors; there are 2 3's in common, so 3*3 = 9 is the GCF. This makes the product 9×( ). The missing number would be 5, since 9*5 = 45. Writing this as a sum, we can use 2+3; this gives us 9×(2+3).

4 0
2 years ago
Read 2 more answers
Which equation has a graph that is a vertical line? A) y+5=3 B) 2x=y C) x-y=0 D) 3x-2=0
aleksandr82 [10.1K]

Answer:

D

Step-by-step explanation:

the equation of a vertical line is of the form x = c where c is a constant

3x - 2 = 0 is in this form if we add 2 to both sides and divide by 3

3x = 2 ⇒ x = \frac{2}{3} → D



6 0
2 years ago
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