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2 years ago

How to find points of inflection from first derivative graph

1 answer:

7 0

To find where the second derivative of the function changes sign, from negative to positive, or vice-versa. So, we find the second derivative of the given function.

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PLZ HELP ASAP Find f(x) if it is known that f(x−3)=0.2x−0.6.

iren [92.7K]

Answer:

f(x) = 0.2x

Step-by-step explanation:

hello :

let : f(x) = ax+b calculate a and b

f(x-3) =a(x-3)+b = ax+b-3a.....(*)

given : f(x-3) = 0.2x-0.6...(**)

by (*) and(**) : a= 0.2 and b-3a = - 0.6

b-3(0.2) = - 0.6

so : b=0

conclusion : f(x) = 0.2x

4 0

3 years ago

4x + y = 3

cestrela7 [59]

Answer:

-1

Step-by-step explanation:

You have to use simultaneous linear equation as there are 2 equations given

use the subtitution method as what I've shown above

Hope you can understand well

Stay safe!

4 0

3 years ago

Read 2 more answers

Simplify the expression (4x^3y^-5)^4

Elanso [62]

Answer:

64x^12 / y^20

Step-by-step explanation:

(4x^3y^-5)^4

Distribute the exponential value of 4 to all terms in the parentheses.

4^4 * x^(3*4) * y^(-5*4)

Multiply each exponent.

64x^12 * y^-20

Make the negative exponent positive by making it a fraction.

64x^12 / y^20

5 0

3 years ago

- Simplify the expression: -21 - (38)

Nutka1998 [239]

Step-by-step explanation:

-21 - (38)

-21-38

-(21+38)

-59

6 0

3 years ago

Read 2 more answers

BRAINLIEST ASAP! PLEASE HELP ME :)

Nataly_w [17]

Step-by-step explanation:

$\displaystyle y = Asin(Bx - C) + D$

According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:

$\displaystyle Phase\:[Horisontal]\:Shift → \frac{\frac{π}{3}}{3} = \frac{π}{9} \\ Period → \frac{2}{3}π$

Therefore we have our answer.

Extended Information on the trigonometric function

$\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|$

NOTE: Sometimes, your vertical shift might tell you to shift your graph below or above the midline where the amplitude is.

I am joyous to assist you anytime.

5 0

2 years ago