# Need help! Minimum Problem

• Oct 14th 2009, 10:57 PM
Zippyz
Need help! Minimum Problem
Hi (Hi)

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. I tried solving it using the quadratic formula for the hell of it. 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, and I have a feeling that it's something really stupid. Any help is appreciated. Thank you. (Rock)
• Oct 14th 2009, 11:22 PM
Prove It
Quote:

Originally Posted by Zippyz
Hi (Hi)

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. I tried solving it using the quadratic formula for the hell of it. 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, and I have a feeling that it's something really stupid. Any help is appreciated. Thank you. (Rock)

I can't think of a way to answer this question without using calculus...

With calculus...

$\displaystyle y = 9t^4 + 6t^2 + 2$

$\displaystyle \frac{dy}{dt} = 36t^3 + 12t$

$\displaystyle 0 = 36t^3 + 12t$

$\displaystyle 0 = 12t(3t^2 + 1)$

$\displaystyle 12t = 0$ or $\displaystyle 3t^2 + 1 = 0$.

Clearly the only possible solution is $\displaystyle t = 0$.

Now to check that it is a minimum...

$\displaystyle \frac{d^2y}{dt^2} = 108t^2 + 12$

At $\displaystyle t = 0, \frac{d^2y}{dt^2} = 12 > 0$.

Thus there is a minimum at $\displaystyle t = 0$.

At $\displaystyle t = 0, y = 2$.

So the minimum is $\displaystyle (t, y) = (0, 2)$.
• Oct 14th 2009, 11:28 PM
Zippyz
Quote:

Originally Posted by Prove It
I can't think of a way to answer this question without using calculus...

With calculus...

$\displaystyle y = 9t^4 + 6t^2 + 2$

$\displaystyle \frac{dy}{dt} = 36t^3 + 12t$

$\displaystyle 0 = 36t^3 + 12t$

$\displaystyle 0 = 12t(3t^2 + 1)$

$\displaystyle 12t = 0$ or $\displaystyle 3t^2 + 1 = 0$.

Clearly the only possible solution is $\displaystyle t = 0$.

Now to check that it is a minimum...

$\displaystyle \frac{d^2y}{dt^2} = 108t^2 + 12$

At $\displaystyle t = 0, \frac{d^2y}{dt^2} = 12 > 0$.

Thus there is a minimum at $\displaystyle t = 0$.

At $\displaystyle t = 0, y = 2$.

So the minimum is $\displaystyle (t, y) = (0, 2)$.

thank you for your reply. but there's no way this can be the only way to do it (unless the book is wrong, which it actually is in some cases believe it or not)

this is way over MY head as well as way over anything in the text book. what you posted above is completely foreign.

The only tools that this class has provided is the vertex formula and completing the square for minimums. And also changing equations that are closely related to a quadratic into a quadratic. For example x^4 + x^2 + c into t^2 + t + c. Then setting what you find in the quadratic equation equal to the x^2 in the original. For example t=x^2. 3=x^2. root 3=x.
• Oct 14th 2009, 11:41 PM
mr fantastic
Quote:

Originally Posted by Zippyz
Hi (Hi)

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. I tried solving it using the quadratic formula for the hell of it. 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, and I have a feeling that it's something really stupid. Any help is appreciated. Thank you. (Rock)

$\displaystyle (3t^2 + 1)^2 + 1$ is a minimum when $\displaystyle 3t^2 + 1$ is a minimum. $\displaystyle 3t^2 + 1$ is a minimum when $\displaystyle t = 0$.
• Oct 14th 2009, 11:52 PM
Zippyz
Quote:

Originally Posted by mr fantastic
$\displaystyle (3t^2 + 1)^2 + 1$ is a minimum when $\displaystyle 3t^2 + 1$ is a minimum. $\displaystyle 3t^2 + 1$ is a minimum when $\displaystyle t = 0$.

can you please explain to me how you came to this conclusion? I really appreciate it (Bow)
• Oct 15th 2009, 12:01 AM
Prove It
Quote:

Originally Posted by Zippyz
can you please explain to me how you came to this conclusion? I really appreciate it (Bow)

Mr F has noticed that

$\displaystyle y = 9t^4 + 6t^2 + 2$

$\displaystyle = 9t^4 + 6t^2 + 1 + 1$

$\displaystyle = (3t^2 + 1)^2 + 1$

$\displaystyle = x^2 + 1$ if $\displaystyle x = 3t^2 + 1$.

So the minimum value for $\displaystyle y$ will occur where $\displaystyle x$ is at its minimum.

So in other words, we need to minimise $\displaystyle 3t^2 + 1$.

I think you can see that that happens where $\displaystyle t= 0$.