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3 years ago

Use the graph of the function f to determine the given limit. Picture provided below

Mathematics

2 answers:

slamgirl [31]3 years ago

7 0

Answer:

a. 2

Step-by-step explanation:

Here we have to find when lim x ---> 2, what would be the value of f(x).

The value of f(x) is nothing y value in the graph.

From the given graph, we have to see the value of y, when x = 2

The value of y = 2, when x = 2 by looking at the graph.

Therefore, the answer is a. 2

Hope you will understand the concept.

Thank you.

elena-14-01-66 [18.8K]3 years ago

6 0

Answer:

Choice B..... The limit seems to be 8

Step-by-step explanation:

From the graph, the value of the function when x approaches 2 would be 8. The point (2, 2) is an undefined point or a point of discontinuity. Therefore, the limit can not be equal to 2. Nevertheless, there is a point marked by a dark circle (2, 8). this means that when x is 2 the value of the function is 8 even though the point does not lie on the straight line.

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Write the equation of the line MN, which is formed by the

krek1111 [17]

Answer:

-1

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find a function where f(0)=2 and f(1)=2

xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

$f(x) = x^2-x+2$

Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$