The graph of the function y=x^3+12x^2+15x+3 has a relative maximum at x=

Set the derivative equal to zero.

y' = 3x^2 + 24x +15 = 0
x^2 +8x +5 = 0
x = (1/2)[-8 + sqrt34]
or
x = (1/2)[-8 - sqrt34]

There are two solutions. Pick the one for which the second derivative is negative.

Only one of them works.

cgcjc

To find the relative maximum of the function, we need to determine the critical points. The critical points are the values of x where the derivative of the function is equal to zero or does not exist.

Let's find the derivative of the function first:
y = x^3 + 12x^2 + 15x + 3

Taking the derivative with respect to x:
dy/dx = 3x^2 + 24x + 15

Now, set the derivative equal to zero and solve for x:
3x^2 + 24x + 15 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. Factoring is not possible in this case, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 3, b = 24, and c = 15. Plugging these values into the quadratic formula, we get:
x = (-24 ± √(24^2 - 4*3*15)) / (2*3)
x = (-24 ± √(576 - 180)) / 6
x = (-24 ± √396) / 6
x = (-24 ± √(4 * 99)) / 6
x = (-24 ± 2√99) / 6
x = -4 ± (√99 / 3)

So, the critical points are:
x = (-4 + √99) / 3
x = (-4 - √99) / 3

To determine whether each critical point is a relative maximum or minimum, we need to analyze the second derivative. If the second derivative is positive, the critical point is a relative minimum, and if it is negative, the critical point is a relative maximum.

Taking the derivative of the first derivative:
d²y/dx² = 6x + 24

Now, substitute the value of x:
d²y/dx² at x = (-4 + √99) / 3 = 6((-4 + √99) / 3) + 24
= -8 + 2√99 + 24
= 16 + 2√99

d²y/dx² at x = (-4 - √99) / 3 = 6((-4 - √99) / 3) + 24
= -8 - 2√99 + 24
= 16 - 2√99

Since both values are positive, we can conclude that there is no relative maximum in the given function.

Therefore, the graph of the function y = x^3 + 12x^2 + 15x + 3 does not have a relative maximum.

To find the relative maximum of a function, we need to find the critical points. Remember that critical points occur where the derivative of the function is either zero or undefined.

Let's first find the derivative of the function y=x^3+12x^2+15x+3. The power rule can be used for differentiating terms with x raised to a power.

dy/dx = 3x^2 + 24x + 15

To find the critical points, we need to solve the equation 3x^2 + 24x + 15 = 0. This is a quadratic equation, and it can be solved using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Applying the quadratic formula to our equation, where a = 3, b = 24, and c = 15:

x = (-24 ± √(24^2 - 4*3*15)) / (2*3)

Simplifying further:

x = (-24 ± √(576 - 180)) / 6
x = (-24 ± √396) / 6
x = (-24 ± 2√99) / 6
x = -4 ± √99/3

So we have two possible solutions for x: x = -4 + √99/3 and x = -4 - √99/3.

To determine which of these values correspond to a relative maximum, we need to analyze the function's behavior around these points. We can do this by looking at the second derivative. If the second derivative is negative at a point, it indicates a relative maximum.

The second derivative of our original function is:

d^2y/dx^2 = 6x + 24

Now substitute the possible x-values we found:

For x = -4 + √99/3:
d^2y/dx^2 = 6(-4 + √99/3) + 24

Simplifying:
d^2y/dx^2 = -24 + 2√99 + 24
d^2y/dx^2 = 2√99

Since the second derivative is positive for this value of x, it does not correspond to a relative maximum.

For x = -4 - √99/3:
d^2y/dx^2 = 6(-4 - √99/3) + 24

Simplifying:
d^2y/dx^2 = -24 - 2√99 + 24
d^2y/dx^2 = -2√99

Since the second derivative is negative for this value of x, it corresponds to a relative maximum.

Therefore, the graph of the function y=x^3+12x^2+15x+3 has a relative maximum at x = -4 - √99/3.