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lisov135 [29]
3 years ago
15

How are the functions y = x and y = x + 2 related? How are their graphs related?

Mathematics
1 answer:
Andru [333]3 years ago
3 0

Answer:

y is more than x and one a graph it is translated up 2 units

Step-by-step explanation:

hope this helps, have a great day.

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Find the area of the shape shown below.
vovangra [49]

Answer:

I believe a=22

Step-by-step explanation:

a=4+4+4+2+8

a=8+4+2+8

a=12+2+8

a=14+8

a=22

8 0
2 years ago
Consider functions of the form f(x)=a^x for various values of a. In particular, choose a sequence of values of a that converges
sleet_krkn [62]

Answer:

A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.

Step-by-step explanation:

A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):

f(x)=a^{x}\\ f'(x)=a^xln(a)

The process will help us to understand what is happening, at first we rewrite the function:

f(x)=a^x\\ f(x)=e^{ln(a^x)}\\ f(x)=e^{xln(a)}\\

And then, we use the chain rule to differentiate:

f'(x)=e^{xln(a)}ln(a)\\ f'(x)=a^xln(a)

Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know  that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).

In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):

if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b)  (and below we explain why this happen)

Considering that ln(a) is a growing function and ln(e)=1, we have:

if a<e<b, then ln(a)< 1 <ln(b)

if a<e, then aˣln(a)<aˣ

if e<b, then bˣ<bˣln(b)

And because eˣ is defined to be the same as its derivative, the cases above results in the following

if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)

if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )

if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )

but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)

3 0
3 years ago
Can someone help me find the relative maximum and minimum? please i really need help
bearhunter [10]

9514 1404 393

Answer:

  • relative maximum: -4
  • relative (and absolute) minimum: -5

Step-by-step explanation:

The curve has a relative maximum where values on either side are lower. This looks like a peak in the curve. There is one of those on the y-axis at y = -4.

The relative maximum is -4.

__

A relative minimum is a low point, where the curve is higher on either side. There are two of these, located symmetrically about the y-axis. The minimum appears to be about y = -5. (They might be at x = ± 1, but it is hard to tell.)

The relative minima are -5.

__

A minimum or maximum is absolute if no part of the curve is lower or higher. Here, the minima are absolute, while the maximum is only relative. (The left and right branches of the curve go higher than y=-4.)

_____

Identifying the points on the curve should be the easy part. Deciding what the coordinates are can be harder when the graph is like this one.

4 0
2 years ago
Find the distance between each pair of points.<br> (3,5) (2,-5)
Sholpan [36]
The distance between these points is (4,7) and (3,-7)
3 0
3 years ago
Hi ! How to solve this question?
tester [92]

You should actually have

-x^2-4x-10=-(x^2+4x+10)=-(x^2+4x+4+6)=-((x+2)^2+6)=-(x+2)^2-6

Now, remember that x^2 is always non-negative, so (x+2)^2\ge0 for any value of x. This means -(x+2)^2\le0 for any x, and so

-(x+2)^2-6\le0-6=-6

i.e. f(x) is at most -6, and hence negative for all x.

4 0
3 years ago
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