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likoan [24]
2 years ago
7

Find each x-value at which f is discontinuous and for each x-value, determine whether f is continuous from the right, or from th

e left, or neither.
f(x) = x + 9 if x < 0
e^x if 0 ≤ x ≤ 1
4 − x if x > 1
x = (smaller value)
continuous from the right, or continuous from the left, or neither
x = (larger value)
continuous from the right, or continuous from the left, or neither
Mathematics
1 answer:
fredd [130]2 years ago
7 0

Using continuity concepts, the answers are given by:

  • The function is right-continuous at x = 0.
  • The function is left-continuous at x = 1.

------------------------------------

A function f(x) is continuous at x = a if:

\lim_{x \rightarrow a^{-}} f(x) = \lim_{x \rightarrow a^{+}} f(x) = f(a)

  • If \lim_{x \rightarrow a^{-}} f(x) = f(a), the function is left-continuous.
  • If \lim_{x \rightarrow a^{+}} f(x) = f(a), the function is right-continuous.
  • For the given function, the continuity is tested at the points in which the definitions are changed, x = 0 and x = 1.

------------------------------------

At x = 0.

  • Approaching from the left, it is less than 0, thus:

\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0} x + 9 = 0 + 9 = 9

  • Approaching from the right, it is more than 0, thus:

\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0} e^x = e^0 = 1

  • The numeric value is:

f(x) = e^0 = 1

  • Since [tex]\lim_{x \rightarrow 0^+} f(x) = f(0), the function is right-continuous at x = 0.

------------------------------------

At x = 1.

\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1} e^x = e^1 = e

\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1} 4 - x = 4 - x = 3

f(1) = e^1 = e

  • Since [tex]\lim_{x \rightarrow 1^-} f(x) = f(1), the function is left-continuous at x = 0.

A similar problem is given at brainly.com/question/21447009

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HELP ME PLEASEE
Nina [5.8K]

Answers:

x = -8/5 or x = 8/5

Sum of the first ten terms where all terms are positive = 4092

========================================================

Explanation:

r = common ratio

  • first term = 4
  • second term = (first term)*(common ratio) = 4r
  • third term = (second term)*(common ratio) = (4r)*r = 4r^2

The first three terms are: 4, 4r, 4r^2

We're given that the sequence is: 4, 5x, 16

Therefore, we have these two equations

  • 5x = 4r
  • 4r^2 = 16

Solve the second equation for r and you should find that r = -2 or r = 2 are the only possible solutions. If r = -2, then 5x = 4r solves to x = -8/5. If r = 2, then 5x = 4r solves to x = 8/5.

-----------------

To find the sum of the first n terms, we use this geometric series formula

Sn = a*(1 - r^n)/(1 - r)

We have

  • a = 4 = first term
  • r = 2, since we want all the terms to be positive
  • n = 10 = number of terms to sum up

So,

Sn = a*(1 - r^n)/(1 - r)

S10 = 4*(1 - 2^10)/(1 - 2)

S10 = 4*(1 - 1024)/(-1)

S10 = 4*(-1023)/(-1)

S10 = 4092

8 0
2 years ago
What is 3/8 x 3/8 x 3/8 written as a power
julia-pushkina [17]

Answer:

3/8^3

Step-by-step explanation:

its like 1/2 x 1/2 x1/2 its 1/2^3

or  7x7x7x7x7=7^5 its the same concept....

7 0
9 months ago
Write an equation fo the line that passrs through (4,-5) and (3,-2)
kap26 [50]
D is the correct answer luv
7 0
3 years ago
The prize of bronze has increased by 10% per year from 2000. In the year 2000, Harry bought a bronze medal for $120. Which of th
barxatty [35]
Hey there!

Consider the exponential growth equation:

f(x) = P(1 + r)^{t}, where:

P = initial amount
r = interest rate (as a decimal, divide a percentage by 100 to get its decimal form)
t = years

All you need to do is plug in your given numbers:

P = 120
r = 0.10
t = x

Your final equation will be:

f(x) = 120(1.10)^{x}

Hope this helped you out! :-)
6 0
3 years ago
Read 2 more answers
Need help please <br><br>what is the minimum value of this function?​
Ksju [112]
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