Question

dy/dx= x^2-2x-4, y(3)=-6, then 3y is equal to
a.) x^3+3x^2+12x-18
b.) x^3-3x^2+12x+18
c.) x^3+3x^2+12x+18
d.) x^3-3x^2-12x+18​

Answer 1

Answer:

The first step to find the min/max values is to find the derivative.

y=2x3-3x2-12x+18

y'=6x2-6x-12

Then we find the x intercepts of the derivative.

0=6x2-6x-12

0=6(x2-x-2)

0=6(x-2)(x+1)

x=2,-1

The values for the min/max are x=-1 and x=2. This is because these are the values at which the derivative is 0. Since the derivative tells the instantaneous slope of the function at any point, 0 values indicate that the slope is changing from positive to negative, which occurs at mix/max values.

We solve for these points in the original function:

y(-1)=-2-3+12+18=31 (-1,25)

y(2)=16-12-24+18=-10 (2,-10)

However to determine which is min and which is max, we need the 2nd derivative.

y'=6x2-6x-12

y''=12x-6

Then we plug in the values we found in the last step.

y''(-1)=12(-1)-6=-18

y''(2)=12(2)-6=18

This is all the info we needed. Since y''(-1) is negative this tells us that this point is a maximum, since a negative 2nd derivative tells us that the slope is decreasing.

Y''(2) is positive, so this is a minimum value since this shows that the slope is increasing.

Bottom line:

Min (2,-10)

Max (-1,25)

Answer 2

Answer:

option d)

Step-by-step explanation:

Given :

$\frac{dy}{ydx} = x^2-2x-4$

$y(3)=-6$

then,

find value of 3y,.

Solution :

We know that,

$\frac{dy}{dx} = x^2-2x-4$

By taking integration,

We get,

⇒ $y(x) = \frac{x^{2+1}}{2+1}-2(\frac{x^{1+1}}{1+1})-4(\frac{x^{0+1}}{0+1}) + C$

⇒ $y(x)=\frac{x^{3}}{3}-2(\frac{x^{2}{2})-4(\frac{x^{1}}{1}) + C$

⇒ $y(x)=\frac{x^{3}}{3}-x^{2}-4x + C$

As, $y(3)=-6$

⇒ $-6 = \frac{(3)^3}{3} - (3)^2 -4(3) +C$

⇒ $-6 = 9 - 9 - 12 +C$

⇒ $-6 = -12 + C$

⇒  $C = 6$

∴ $3y = 3(\frac{x^{3}}{3}-x^{2}-4x +6)$

⇒ $3y = x^{3}-3x^{2}-12x +18$

∴ Option d) is correct.