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Gary B.

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Calculus

TutorMe

Question:

Find the critical points for the equation y = 5x^3 + 2x^2 -6x + 1 and give the concavity of each point.

Gary B.

Answer:

We will need to find the first derivative to acquire the critical points, which is y' = 15x^2 +4x - 6. The critical points are acquired by setting the first derivative equal to zero and solving for x... 0 = 15x^2 +4x - 6 x = 0.513..., x = -0.779... After finding the critical x-values, we will plug these x-values into the original equation to get corresponding y-values, which are (-0.779, 4.524) and (0.513, -0.877). We will need the second derivative in order to determine the concavity of these points. The second derivative is y" = 30x + 4. By plugging in the x-values, we will find that (-0.779, 4.524) is concave down and (0.513, -0.877) is concave up.

Algebra

TutorMe

Question:

Will the parabola formed by the equation y = x^2 + 5x + 6 touch the x-axis? Why or why not?

Gary B.

Answer:

The given equation factors into y = (x + 2)(X + 3), which gives x = -3 and x = -2 as roots. Therefore, the parabola produced by the equation y = x^2 + 5x + 6 will indeed touch the x-axis.

Differential Equations

TutorMe

Question:

Find the particular solution of xy/(x + 1) = dy/dx with the initial value of y(0) = 1.

Gary B.

Answer:

The first order of business will be to find the general solution. This is a separation of variables problem because it takes the form of dy/dx = f(x)*g(y). We will follow the following procedure. STEP ONE: Multiply both sides by dx... xy/(x+1) dx = dy STEP TWO: Divide both sides by y... x/(x+1) dx = 1/y dy STEP THREE: Integrate Both Sides... ∫x/(x+1) dx = ∫1/y dy x − ln(|x+|) = ln|y| + C STEP FOUR: Find "C" by using the initial value x − ln(|x+|) = ln|y| + C 0 − ln(|0+|) = ln|1| + C 0 - 0 = 0 + C C= 0 STEP FIVE: Write the exact solution x − ln(|x+|) = ln|y|

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