Math (Pre-Calculus)

posted by .

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.

  • Math (Pre-Calculus) -

    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.

  • Math (Pre-Calculus) -corr. -

    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Pre-calculus-check answers

    Name all the values of x that are not in the domain of f(x)=2-x^2/x+5. Answer: x= -5 2)Find the minimum value of f(x,y)=2x-y+2 for the polygonal convex set determined by this system of inequalities: x >/= 1, x </= 3, y</=0, …
  2. Pre-calculus

    Name all the values of x that are not in the domain of f(x)=2-x^2/x+5. Answer: x= -5 2)Find the minimum value of f(x,y)=2x-y+2 for the polygonal convex set determined by this system of inequalities: x >/= 1, x </= 3, y</=0, …
  3. Pre-Calculus-check answers

    Name all the values of x that are not in the domain of f(x)=2-x^2/x+5. Answer: x= -5 2)Find the minimum value of f(x,y)=2x-y+2 for the polygonal convex set determined by this system of inequalities: x >/= 1, x </= 3, y</=0, …
  4. Math

    The question is: Fr what value(s) of k will the function f(x)=(kx^2)-4x+k have no zeroes. So for that, the discriminant would be less than 0. 0>D 0>((-4)^2)-4(k)(k) 0>16-(4k^2) 4k^2>16 k^2>4 k>+2/-2 So I know that …
  5. pre-calculus

    Let f(x)=x^2+bx+27. Find the value(s) of b so that f(x) has minimum value of 2.
  6. Pre-calculus

    Find the value for c for which the system is consistent and dependent. 3x + y = 5 4.5y = c + 9x Find the minimum value of f(x, y) = y – x + 1 for the polygonal convex set determined by this system of inequalities. x > 0 y > …
  7. Pre-calculus

    Find the minimum value of f(x, y) = y – x + 1 for the polygonal convex set determined by this system of inequalities. x > 0 y > 0 2x + y< 4 the answer is -1. please tell me why
  8. Pre-Calculus

    Hello! I'm stuck on how to do this problem and would appreciate some help! Choose a non-zero complex number c for f(z) = z2+ c. Find a complex number v that is in the escape set. Find a complex number w that is in the prisoner set. …
  9. Pre-Calculus

    Determine a quadratic function in vertex form given each set of characteristics. * minimum value of -24 and x-intercepts at -21 and -5 I have: (-21,0) and (-5,0) How would I find the x-coordinate of the vertex?
  10. Calculus

    Find the minimum value of the following function: h(x)= x-(12(x+1)^(1/3) -12) on the interval [0, 26] Honestly, I have found by graphing that the minimum value is at the coordinate point (9,-3), but I need to do this in a way that …

More Similar Questions