If t is a constant and the curve y=2x^3+3tx^2+6x+131 has NO horizontal tangents, what conditions must t satisfy?

(a) t >-2
(b) t < 2
(c) t > 2
(d) t < -2
(e) none of these

Apparently, the correct answer is (e).

So for the curve to have no horizontal tangents, I know it's derivative cannot equal to 0.

If y=2x^3+3tx^2+6x+131, then
y'=6x^2+6tx+6 cannot equal 0
y'=6(x^2+tx+1) cannot equal 0

What are the following steps that what would bring me to answer (e) none of these?

6(x^2+tx+1) cannot equal 0

so, the discriminant (b^2-4ac) must be negative
t^2 - 4 < 0
|t| < 2

Your explanation makes sense. However, if I graph f(x)=6(x^2+x+1), for example, on desmos I can see there is a horizontal tangent line at (-0.5, 4.5). This observation contradicts your answer since |1| < 2. Why is this?

6(x^2+x+1) is the derivative. It does not matter where it has a horizontal tangent.

We want to find where y=2x^3+3tx^2+6x+131 has a horizontal tangent. It does not, because 6(x^2+x+1) is never zero.

To determine the conditions that the constant t must satisfy for the curve to have no horizontal tangents, you need to find the values of t for which the derivative of the curve, y', is not equal to zero. Let's break down the steps to find the answer:

Step 1: Find the derivative of y with respect to x, denoted as y'.

Given y = 2x^3 + 3tx^2 + 6x + 131, we differentiate each term with respect to x using the power rule. Since the derivative of a constant term (131 in this case) is zero, we can ignore it in the differentiation process. Thus, we focus on the polynomial terms only:

dy/dx = d/dx(2x^3) + d/dx(3tx^2) + d/dx(6x)

Step 2: Simplify the derivative expression.

Using the power rule, we differentiate each term with respect to x:

dy/dx = 6x^2 + 6t*(2x) + 6

Simplifying further, we have:

dy/dx = 6x^2 + 12tx + 6

Step 3: Set the derivative dy/dx equal to zero and solve for x.

To find the x-values at which the slope of the curve is zero (i.e., horizontal tangents), we need to solve the equation:

6x^2 + 12tx + 6 = 0

Step 4: Determine the discriminant of the quadratic equation.

The discriminant, denoted as D, can be calculated using the formula D = b^2 - 4ac, where the quadratic equation is ax^2 + bx + c = 0.

Here, a = 6, b = 12t, and c = 6. Thus, the discriminant is:

D = (12t)^2 - 4(6)(6) = 144t^2 - 144

Step 5: Determine the conditions for t to satisfy to have no horizontal tangents.

For the curve to have no horizontal tangents, the discriminant (D) must be greater than zero. That is, D > 0.

Therefore, we have:

144t^2 - 144 > 0

Dividing both sides by 144, we get:

t^2 - 1 > 0

(t - 1)(t + 1) > 0

To solve this inequality, we consider the sign of the expression (t - 1)(t + 1):

When t < -1 or t > 1, both factors are negative, producing a positive result.

When -1 < t < 1, the first factor is negative, but the second factor is positive, resulting in a negative result.

Step 6: Determine the conditions for t.

From the above analysis, we see that the inequality t^2 - 1 > 0 holds when either t < -1 or t > 1.

However, none of the answer choices (a) through (d) correspond to this result. Therefore, the correct answer is (e) none of these.

In conclusion, for the curve y = 2x^3 + 3tx^2 + 6x + 131 to have no horizontal tangents, the value of t must fall into the range t < -1 or t > 1.