9t^4 + 6t^2 + 2

the problem is asking for the value of t at which the graph is at it's minimum.

what i did was set t^4 to x^2 and t^2 to x. So that means I have 9x^2 + 6x + 2.

Completing the square and the vertex formula gave me the same answer of -1/3 as x. So setting that equal to t^2 and solving gives me a no solution answer. The same thing happens when I try to solve it using the quad formula. 36-72 shows up in the radical, which is negative.

So I'm at a loss, my answer comes out to be undefined.

But the answer to the problem (as stated in the back of the book) is x=0. Also when I put it in a graphing calculator the minimum is indeed 2 at t=0.

As far as I know, undefined is not the same as 0. I must be doing something wrong. Any help is appreciated. Thank you.

If it is precalculus, you're not expected to use derivatives, which incidentally will give you x=0 as the answer.

I do not know if you teacher has shown you a systematic method to calculate the minimum for a quartic. In this particular situation, we can find the answer rapidly by taking advantage of the way the function is made up.

f(x) = 9x⁴ + 6x² + 2
is made up of three terms, two of which are non-negative. So the minimum is 2 when both of the non-negative terms vanish. It is obvious that 9x⁴ and 6x² vanish at x=0. The vertex is therefore (0,2).

I also note that f(x) can be composed by goh(x) where g(x)=9x²+6x+2 and h(x)=x². Not that this property will help us find the vertex without the use of calculus, as far as I know.

Also, I have not come across a method to find the vertex for a general quartic without the use of calculus. If anyone would share the answer, it would be appreciated.

last paragraph: `"the vertices for a general quartic ".

Also, I have not mentioned that graphing by tabulation almost always works for finding extrema of polynomials, although a good judgment goes a long way.

To find the value of t at which the graph is at its minimum, we need to find the minimum value of the function. In this case, the function is 9t^4 + 6t^2 + 2.

To proceed, we can rewrite the function as 9t^4 + 6t^2 + 2 = (3t^2)^2 + 2(3t^2) + 2.

Now, let's rewrite the expression as a quadratic equation: y = x^2 + 2x + 2, where x = 3t^2.

To find the minimum value of this quadratic equation, we can use the vertex formula. The vertex formula states that for a quadratic equation in the form y = ax^2 + bx + c, the x-coordinate of the vertex can be found using the formula x = -b / (2a).

In our case, a = 1 and b = 2. So, the x-coordinate of the vertex is x = -2 / (2*1) = -1.

Now, we need to convert x back into t. Recall that we defined x = 3t^2. So, -1 = 3t^2.

Dividing both sides of the equation by 3 gives: -1/3 = t^2.

Taking the square root of both sides gives: t = ±√(-1/3).

At this point, we encounter a problem. The square root of a negative number is not defined in the realm of real numbers. Hence, we do not have a real solution for t.

However, let's consider the original equation without replacing t^4 with x^2 and t^2 with x. In this case, we have 9t^4 + 6t^2 + 2.

If we graph this equation, we will find that the graph is symmetric and has a minimum point. The minimum occurs at t = 0, which matches the answer given in the book and the result obtained from a graphing calculator.

The issue with using the vertex formula and completing the square arises from trying to solve for t by converting the equation into a quadratic form. In this particular case, it leads to an undefined result because of the presence of a negative inside the square root. However, simply analyzing the original equation without substituting variables gives us the correct answer.

To find the value of t at which the graph is at its minimum, we can use calculus. The minimum point of a function occurs at the derivative equals zero (assuming it exists) or at the endpoints of the interval under consideration.

First, let's find the derivative of the function 9t^4 + 6t^2 + 2 with respect to t. The derivative is obtained by applying the power rule:

d/dt(9t^4 + 6t^2 + 2) = 36t^3 + 12t

Now, set this derivative equal to zero and solve for t:

36t^3 + 12t = 0

Factor out t:

t(36t^2 + 12) = 0

Either t = 0 or (36t^2 + 12) = 0.

Setting (36t^2 + 12) = 0:

36t^2 + 12 = 0

Divide through by 12:

3t^2 + 1 = 0

Rearrange the terms:

3t^2 = -1

Divide through by 3:

t^2 = -1/3

This equation has no real solutions since the square of a real number cannot be negative. Therefore, t=0 is the only possible solution.

Now, let's verify that t=0 indeed corresponds to the minimum of the graph. We can either use the second derivative test or the fact that the leading coefficient is positive to confirm it.

The second derivative is obtained by differentiating the derivative we found earlier:

d^2/dt^2(9t^4 + 6t^2 + 2) = 108t^2 + 12

To apply the second derivative test, we substitute t=0 into the second derivative:

108(0)^2 + 12 = 12

Since the second derivative is positive, it confirms that t=0 corresponds to a local minimum.

Hence, the value of t at which the graph is at its minimum is t=0.