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Reika [66]
3 years ago
6

Explain what the vertical line test is and how it is used.

Mathematics
2 answers:
topjm [15]3 years ago
5 0

Answer:

The vertical line test is a way to determine if a relation is a function. This test determines if one input has exactly one output on the graph. If any vertical line passes through more than one point on the graph, then the relation is not a function because two different outputs have the same input

Likurg_2 [28]3 years ago
4 0

Answer:

The vertical line test is  is a visual way to determine if a curve is a graph of a function or not.

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5 times

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What would be the total cost of Annabel renting the surfboard for 6 hours
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Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = e−4x, [0, 2] Yes, it does not m
Zigmanuir [339]

Answer:

(a) Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem.

(b) c =0.51995

Step-by-step explanation:

Given

f(x) = e^{-4x};\ [0,2]

Solving (a); Does the function satisfy M.V.T on the given interval

We have:

f(x) = e^{-4x};\ [0,2]

The above function is an exponential function, and it is differentiable and continuous everywhere

Solving (b): The value of c

To do this, we use:

f'(c) = \frac{f(b) - f(a)}{b - a}

In this case:

[a,b] = [0,2]

So, we have:

f'(c) = \frac{f(2) - f(0)}{2 - 0}

f'(c) = \frac{f(2) - f(0)}{2}

Calculate f(2) and f(0)

f(x) = e^{-4x}

So:

f(2) = e^{-4*2} = e^{-8} = 0.00033546262

f(0) = e^{-4*0} = e^{0} = 1

This gives:

f'(c) = \frac{0.00033546262 - 1}{2}

f'(c) = \frac{-0.99966453738}{2}

f'(c) = -0.4998

Note that:

f'(x) = (e^{-4x})'

f'(x) = -4e^{-4x}

This implies that:

f'(c) = -4e^{-4c}

So, we have:

f'(c) = -0.4998

-4e^{-4c} =-0.4998

Divide both sides by -4

e^{-4c} =\frac{-0.4998}{-4}

e^{-4c} =0.12495

Take natural logarithm of both sides

\ln(e^{-4c}) =\ln(0.12495)

\ln(e^{-4c}) =-2.0798

Apply law of natural logarithm

\ln(e^{ax}) =ax

So:

-4c =-2.0798

Solve for c

c =\frac{-2.0798}{-4}

c =0.51995

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