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3 years ago

Helpppp Plzzzz ASAPppppppppp

Mathematics

1 answer:

Nataly [62]3 years ago

6 0

Non linear:

y = 6 -x^2

y = 2x^2 + 2

y = x^3

Rate of change is also the slope

= (72 - 69) / (51 - 48)

= 3/3

= 1

Rate of change = 1

First you need to find slope

(3,0) and (5,2)

Slope = (2 - 0)/(5-3) =2/2 = 1

Equation

y - 0 = 1(x - 3)

y = x - 3

Answer:

Tyler's function is correct as y = x - 3, it matched the above.

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3 years ago

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laiz [17]

Answer:

(a) See below.

(b) x = 0 or x = 1

Step-by-step explanation:

Given rational function:

$f(x)=\dfrac{\ln |x-1|}{x}$

Part (a)

Substitute x = 2 into the given rational function:

$\begin{aligned}\implies f(2) & =\dfrac{\ln |2-1|}{2}\\\\ & =\dfrac{\ln 1}{2}\\\\ & =\dfrac{0}{2}\\\\ & =0\end{aligned}$

Therefore, as the function is defined at x = 2, the function is continuous at x = 2.

Part (b)

Given interval: [-2, 2]

Logs of negative numbers or zero are undefined. As the numerator is the natural log of an absolute value, the numerator is undefined when:

|x - 1| = 0 ⇒ x = 1.

A rational function is undefined when the denominator is equal to zero, so the function f(x) is undefined when x = 0.

So the function is discontinuous at x = 0 or x = 1 on the interval [-2, 2].

Part (c)

x = 1 is a vertical asymptote. As the function exists on both sides of this vertical asymptote, it is an infinite discontinuity. Since the function doesn't approach a particular finite value, the limit does not exist. Therefore, x = 1 is a non-removable discontinuity.

A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal.

$\implies f(0)=\dfrac{\ln |0-1|}{0}=\dfrac{\ln 1}{0}=\dfrac{0}{0}$

Therefore, there is a hole at x = 0.

The removable discontinuity of a function occurs at a point where the graph of a function has a hole in it. Therefore, x = 0 is a removable discontinuity.

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