The local minima of $f\left(x\right)=\ \frac{\left(2x+3\right)^2\left(x\ -2\right)^5}{x^3\left(x-5\right)^2}$ are (x, f(x)) = (-1.5, 0) and (7.980, 609.174)

<h3>How to determine the local minima?</h3>

The function is given as:

$f\left(x\right)=\ \frac{\left(2x+3\right)^2\left(x\ -2\right)^5}{x^3\left(x-5\right)^2}$

See attachment for the graph of the function f(x)

From the attached graph, we have the following minima:

Minimum = (-1.5, 0)

Minimum = (7.980, 609.174)

The above means that, the local minima are

(x, f(x)) = (-1.5, 0) and (7.980, 609.174)

Read more about graphs at:

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7 0

2 years ago

How to solve for W because I honestly don’t understand anything

erastova [34]

we do 2x-1/y=w+2/2z

2x-1/y we multiple each side y

2x-1/y×y=2x-1

then w+2/2z

we said×2z×2z

then we got

w+2

later

2x-1=w+2=

then 1 become another side

which is 1+2=3

then

w=3-2x

I hope it's correct

4 0

3 years ago

Read 2 more answers

PLEASE HELP WITH THIS ONE QUESTION

raketka [301]

Answer:

dont know

Step-by-step explanation:

sorry

8 0

3 years ago

During one hour of television there are two shows on what is the ratio of hours to shows if you watch three hours of television

3241004551 [841]

During one hour of television there are two shows on. That would be the ratio 1:2. We need to find 3:?. Since 1×3 is 3, we multiply the second number by 3 as well. 2×3 is 6. So the ratio would be 3:6.

4 0

3 years ago