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3 years ago

Given the following polynomial find the maximum possible number of turning points

1 answer:

8 0

$f'(x)=6x^5+10x^9+9x^8+10x^9=20x^9+9x^8+6x^5$

The degree of the derivative is 9, so the maximum possible number of turning points is 9.

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3 years ago

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Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent

goblinko [34]

Answer:

$y(x)=x^2+5x$

Step-by-step explanation:

Given: $y'=\sqrt{4y+25}$

Initial value: y(1)=6

Let $y'=\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\sqrt{4y+25}$

Variable separable

$\dfrac{dy}{\sqrt{4y+25}}=dx$

Integrate both sides

$\int \dfrac{dy}{\sqrt{4y+25}}=\int dx$

$\sqrt{4y+25}=2x+C$

Initial condition, y(1)=6

$\sqrt{4\cdot 6+25}=2\cdot 1+C$

$C=5$

Put C into equation

Solution:

$\sqrt{4y+25}=2x+5$

$4y+25=(2x+5)^2$

$y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$

$y(x)=x^2+5x$

Hence, The solution is $y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$ or $y(x)=x^2+5x$

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2 years ago