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PIT_PIT [208]
3 years ago
9

Given no other restrictions, what are the domain and range of the following funtction? f(x)=x^2-2x+2

Mathematics
2 answers:
STALIN [3.7K]3 years ago
8 0

Answer: Domain of the given function is all real numbers.

Range of the given function is {y | y>=1}

Step-by-step explanation:

The given function is a polynomial and the domain of any polynomial is all the real numbers because for any value of x, the function will have some value.

Hence, the domain will be all real numbers.

Range is the set of y values for which the function is defined.

This is a polynomial and to find the range, we will simplify the polynomial.

f(x)=x^2-2x+2\\f(x)=(x-1)^2+1

Therefore, for any value of x, f(x) ≥ 1

Hence, the range becomes {y | y ≥ 1}

jeka943 years ago
8 0
F ( x ) = x² - 2 x + 2
The vertex form:
f ( x ) = ( x² - 2 x + 1 )- 1 + 2 = ( x - 1 )² + 1
Coordinates of the vertex are:  ( 1 , 1 ).
Domain:  x ∈ R
Range:  x ∈ [ 1. + ∞ )
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A circle is growing so that the radius is increasing at the rate of 3 cm/min. How fast is the area of the circle changing at the
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Answer:

The area is growing at a rate of \frac{dA}{dt} =226.2 \,\frac{cm^2}{min}

Step-by-step explanation:

<em>Notice that this problem requires the use of implicit differentiation in related rates (some some calculus concepts to be understood), and not all middle school students cover such.</em>

We identify that the info given on the increasing rate of the circle's radius is 3 \frac{cm}{min} and we identify such as the following differential rate:

\frac{dr}{dt} = 3\,\frac{cm}{min}

Our unknown is the rate at which the area (A) of the circle is growing under these circumstances,that is, we need to find  \frac{dA}{dt}.

So we look into a formula for the area (A) of a circle in terms of its radius (r), so as to have a way of connecting both quantities (A and r):

A=\pi\,r^2

We now apply the derivative operator with respect to time (\frac{d}{dt}) to this equation, and use chain rule as we find the quadratic form of the radius:

\frac{d}{dt} [A=\pi\,r^2]\\\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}

Now we replace the known values of the rate at which the radius is growing ( \frac{dr}{dt} = 3\,\frac{cm}{min}), and also the value of the radius (r = 12 cm) at which we need to find he specific rate of change for the area :

\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}\\\frac{dA}{dt} =\pi\,*2*(12\,cm)*(3\,\frac{cm}{min}) \\\frac{dA}{dt} =226.19467 \,\frac{cm^2}{min}\\

which we can round to one decimal place as:

\frac{dA}{dt} =226.2 \,\frac{cm^2}{min}

4 0
4 years ago
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Please help asap for the area. i got homework to submit in 5 minutes
Luba_88 [7]

Answer:

area = 228 m^2

Step-by-step explanation:

here's your solution

=> length = 20 m. , width = 12 m

=> Area of rectangle = length*width

=> Area of rectangle = 12*20

=> Area of rectangle = 240 m^2

now we need to find area of traingle

=> base = 4 m. , height = 6 m

=> area of traingle = 1/2*base*height

=> area of traingle = 1/2*4*6

=> area of traingle = 12 m^2

=> now , area of figure = 240 m^2 - 12 m^2

=> 228 m^2

hope it helps

8 0
3 years ago
Read 2 more answers
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