<img src="https://tex.z-dn.net/?f=y%3Dx%5E%7B2%7D%20%2B3" id="TexFormula1" title="y=x^{2} +3" alt="y=x^{2} +3" align="absmiddle"

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Pavel [41]

3 years ago

class="latex-formula">
What is the domain and range

Mathematics

1 answer:

Katena32 [7]3 years ago

6 0

Answer:

The domain is (negative infinity, infinity) the range is [3. Infinity) y>=3

Step-by-step explanation:

The domain for all quadratic functions are (Negative infinity, infinity)

The range is the output values of the function 3, to infinity

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lbvjy [14]

Step 1:
Start by putting $\frac{d}{dx}$ in front of each term

$\frac{d}{dx}[y cos x]= \frac{d}{dx}[5x^2]+ \frac{d}{dx}[ 3y^2]$
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Step 2:

Deal with the terms in 'x' and the constant terms
$\frac{d}{dx}[ycosx]= 10x+ \frac{d}{dx} [3y^2]$
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Step 3:

Use the chain rule for the terms in 'y'
$\frac{d}{dx}[ycosx]=10x+6y \frac{dy}{dx}$
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Step 4:

Use the product rule on the term in 'x' and 'y'
$(y) \frac{d}{dx} cos x+(cos x) \frac{d}{dx}y =10x+6y \frac{dy}{dx}
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$y(-siny)+(cosx) \frac{dy}{dx} =10x+6y \frac{dy}{dx}$
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Step 5:

Rearrange to make $\frac{dy}{dx}$ the subject
$-y sin(y)+cos(x) \frac{dy}{dx} =10x+6y \frac{dy}{dx}$
$cos(x) \frac{dy}{dx}-6y \frac{dy}{dx}=10x+y sin(y)$
$[cos(x) - 6y] \frac{dy}{dx}=10x + y sin(y)$
$\frac{dy}{dx}= \frac{10x+ysin(y)}{cos(x)-6y}$ ⇒ Final Answer

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