For what values of a and b is the line 3x+y=b tangent to the curve y=ax^3 when x=–3?

Well, if the line is tangent to the curve, that means the slopes of the line and the curve at the point of tangency should be equal. Let's find the slope of the line first.

The equation of the line is given as 3x + y = b, which we can rewrite as y = -3x + b. To find the slope, we can see that the coefficient of x is -3, so the slope of the line is -3.

Now, let's find the slope of the curve y = ax^3. We can do this by taking the derivative of the equation with respect to x. The derivative of ax^3 is 3ax^2.

Now that we have the slope of the line (-3) and the slope of the curve (3ax^2), we can set them equal to each other:

-3 = 3ax^2

Since we know that x = -3, we can substitute it into the equation:

-3 = 3a(-3)^2

Simplifying the equation gives us:

-3 = 3a(9)
-3 = 27a

Dividing both sides by 27 gives us:

-3/27 = a

Simplifying the fraction gives us:

-1/9 = a

So, the value of a is -1/9.

Now that we have the value of a, we can find the value of b. We can substitute the values of a and x into the equation of the line:

y = -3x + b
y = -3(-3) + b
y = 9 + b

Since we know that the point of tangency is (x, y) = (-3, a*(-3)^3), we can substitute the values into the equation of the curve:

y = ax^3
y = (-1/9)(-3)^3
y = (-1/9)(-27)
y = 3

Now we have two equations for y: y = 9 + b and y = 3. Since these two equations represent the same point of tangency, we can set them equal to each other:

9 + b = 3

Solving for b gives us:

b = 3 - 9
b = -6

So, for the line 3x + y = b to be tangent to the curve y = ax^3 at x = -3, the values of a and b are a = -1/9 and b = -6.

To find the values of a and b for which the line 3x+y=b is tangent to the curve y=ax^3 when x=-3, we need to consider the conditions for a line to be tangent to a curve.

First, we can find the derivative of the curve y=ax^3 with respect to x. Taking the derivative gives us:

dy/dx = 3ax^2

To determine if the line 3x+y=b is tangent to the curve at x=-3, we need to find the value of y when x=-3 in both the line equation and the curve equation.

1. For the line y = -3x + b, substituting x=-3 gives us:

y = -3(-3) + b
y = 9 + b

So the y-coordinate of the point on the line when x=-3 is 9+b.

2. For the curve y = ax^3, substituting x=-3 gives us:

y = a(-3)^3
y = -27a

So the y-coordinate of the point on the curve when x=-3 is -27a.

Since the line and the curve are tangent at the point (x, y), we have:

9 + b = -27a

This equation allows us to solve for the values of a and b that satisfy the tangent condition.

Let's proceed with solving the equation:

9 + b = -27a

If we want the line to be tangent to the curve, the slope of the line (-3) must be equal to the derivative of the curve evaluated at x=-3 (which is 3(-3)^2 = 27). This condition gives us:

-3 = 27a

Simplifying this equation, we have:

a = -1/9

Substituting a=-1/9 into the first equation, we can solve for b:

9 + b = -27(-1/9)
9 + b = 3
b = 3 - 9
b = -6

Therefore, the values of a and b for which the line 3x+y=b is tangent to the curve y=ax^3 at x=-3 are a = -1/9 and b = -6.

To find the values of a and b for which the line is tangent to the curve, we need to determine the slope of both the line and the curve at the point of tangency.

The slope of the line can be found by comparing its equation to the slope-intercept form (y = mx + b), where m represents the slope. From the equation 3x + y = b, we can rewrite it as y = -3x + b, indicating that the slope of the line is -3.

To find the slope of the curve, we differentiate the equation y = ax^3 with respect to x. The derivative of ax^3 is 3ax^2, representing the slope of the curve at any given point.

Substitute x = -3 into the derivative to find the slope of the curve at x = -3:
dy/dx = 3ax^2
dy/dx = 3a(-3)^2
dy/dx = 27a

At the point of tangency, the slope of the line and the slope of the curve are equal. Therefore, we set -3 equal to 27a and solve for a:
-3 = 27a
a = -3/27
a = -1/9

Now that we have determined the value of a, we can substitute it back into the equation for the curve (y = ax^3) to find the corresponding value of y at x = -3:
y = (-1/9)(-3)^3
y = (-1/9)(-27)
y = 3

So, at x = -3, the curve y = ax^3 corresponds to y = 3. To make the line 3x + y = b tangent to the curve at x = -3, the y-coordinate of the point of tangency on the line must be 3.

Substitute y = 3 and x = -3 into the equation of the line to find the corresponding value of b:
3(-3) + y = b
-9 + 3 = b
-6 = b

Therefore, the values of a and b for which the line 3x + y = b is tangent to the curve y = ax^3 at x = -3 are a = -1/9 and b = -6.

Differentiate y=ax^3 gives:dy/dx=3ax^2 from 3x+y=b rearranging y=-3x+b m=-3 therefore 3ax^2=-3 ax^2=-1 since x=-3 a(-3)^2=-1 9a=-1 a=-1/9 (b)when x=-3 y=0 0=-3(-3)+b b=-9