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Stella [2.4K]
3 years ago
6

Let f(x) = |x| for all real numbers x. Write the formula for the function represented by the described

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
6 0

Answer:

y=(\frac{1}{3}f(x-3))-1

Step-by-step explanation:

Please see the picture below.

1. Given the function f(x) = |x|, applying a vertical stretch with scale factor \frac{1}{3}, we have the transformed function:

y=\frac{1}{3}f(x)

2. Applying a translation of 3 units to the right, we have:

y=\frac{1}{3}f(x-3)

3. Finally applying a translation down of 1 unit, we have:

y=(\frac{1}{3}f(x-3))-1

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Please please please help and solve this with steps, much help needed thank you :) 20 points for this!
leonid [27]

Answer:

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

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Step-by-step explanation:

\frac{dy}{dx}y=x^3+2x-5x+3

\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}

  • \mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)

1. \mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:

y'\:y=x^3+2x-5x+3

2. \mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}

yy'\:=x^3-3x+3

  • N\left(y\right)\cdot y'\:=M\left(x\right)
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3. \mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1

4. \mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}

5. \mathrm{Simplify}

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

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