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2 years ago

Can someone help me find the relative maximum and minimum? please i really need help

Mathematics

1 answer:

bearhunter [10]2 years ago

4 0

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.

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2 years ago

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Sever21 [200]

Answer:

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3 years ago

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Nataliya [291]

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5 0

3 years ago

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Twice a number, increased by 3 is the same as five less than four times a number.what is the number?

zzz [600]

Answer:

x = 4, hope this helps!

Step-by-step explanation:

2x + 3 = 4x - 5

-2x -2x

3 = 2x -5

+5 +5

8 = 2x

divide by 2 on both sides

4 = x

switch where the x and 4 is, because teachers seem to like that

x = 4

8 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

tangare [24]

Answer:

<h3>The given expression can be written in the form $q(x)+\frac{r(x)}{b(x)}$ as $(2x^2-x+3)+\frac{0}{x-3}$ where q(x)= $2x^2-x+3$ , r(x)=0 and b(x)=x-3</h3>

Step-by-step explanation:

Given rational expression is $\frac{2x^3-7x^2+6x-9}{x-3}$

<h3>To rewrite the given rational expression in the given form and match their correct expressions:</h3>

Rewrite the given rational expression $\frac{2x^3-7x^2+6x-9}{x-3}$ in the form of $q(x)+\frac{r(x)}{b(x)}$ where q(x) is the quotient, r(x) is the remainder and b(x) is the divisor.

<h3>Now solve the given expression by using Synthetic division, we get</h3>

Since x-3 is a factor for $2x^3-7x^2+6x-9$

<h3>Therefore b(x)=x-3</h3>

3_| 2 -7 6 -9

0 6 -3 9

------------------------------------

2 -1 3 0

------------------------------------

Therefore we have quadratic equation $2x^2-x+3=0$

<h3>q(x)= $2x^2-x+3$ </h3><h3>Remainder is 0</h3><h3>Therefore r(x)=0</h3>

Therefore we can write in the form as $(2x^2-x+3)+\frac{0}{x-3}$

It is in the form of $q(x)+\frac{r(x)}{b(x)}$

<h3>Therefore the given expression can be written in the form $q(x)+\frac{r(x)}{b(x)}$ as $(2x^2-x+3)+\frac{0}{x-3}$ where q(x)= $2x^2-x+3$ , r(x)=0 and b(x)=x-3</h3>

3 0

3 years ago

Read 2 more answers