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2 years ago

Find k so that the following function is continuous: f(x)={kx8x2if0≤x<5if5≤x.

Mathematics

1 answer:

tankabanditka [31]2 years ago

3 0

Check the one-sided limits:

$\displaystyle \lim_{x\to5^-}f(x) = \lim_{x\to5}kx = 5k$

$\displaystyle \lim_{x\to5^+}f(x) = \lim_{x\to5}8x^2 = 200$

If f(x) is to be continuous at x = 5, then these two limits should have the same value, which means

5k = 200

k = 200/5

k = 40

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Complete the table of values

Agata [3.3K]

Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify!

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, $3^{-2} =
\frac{1}{3^{2} }$ .
2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, $(\frac{3}{4}) ^{3} = \frac{ 3^{3} }{4^{3} }$ .
3) The zero exponent rule says that any number raised to zero is 1. For example, $3^{0} = 1$ .


Back to the Problem:
Problem 1
The x-values are in the left column. The title of the right column tells you that the function is $y = 4^{-x}$ . The x-values are:
1) x = 0
Plug this into $y = 4^{-x}$ to find letter a:
$y = 4^{-x}\\
y = 4^{-0}\\
y = 4^{0}\\
y = 1$

2) x = 2
Plug this into $y = 4^{-x}$ to find letter b:
$y = 4^{-x}\\ 
y = 4^{-2}\\ 
y = \frac{1}{4^{2}} \\ 
y= \frac{1}{16}$

3) x = 4
Plug this into $y = 4^{-x}$ to find letter c:
$y = 4^{-x}\\ 
y = 4^{-4}\\ 
y = \frac{1}{4^{4}} \\ 
y= \frac{1}{256}$


Problem 2
The x-values are in the left column. The title of the right column tells you that the function is $y = (\frac{2}{3})^x$ . The x-values are:
1) x = 0
Plug this into $y = (\frac{2}{3})^x$ to find letter d:
$y = (\frac{2}{3})^x\\
y = (\frac{2}{3})^0\\
y = 1$

2) x = 2
Plug this into $y = (\frac{2}{3})^x$ to find letter e:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^2\\ y = \frac{2^2}{3^2}\\
y = \frac{4}{9}$

3) x = 4
Plug this into $y = (\frac{2}{3})^x$ to find letter f:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^4\\ y = \frac{2^4}{3^4}\\ y = \frac{16}{81}$

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Answers:
a = 1
b = $\frac{1}{16}$ 
c = $\frac{1}{256}$
d = 1
e = $\frac{4}{9}$
f = $\frac{16}{81}$

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Answer:

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Answer:

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Step-by-step explanation:

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