finding minimum value

**mark** · September 2nd 2009, 05:01 AM

can anyone show me (in detail) how to find the mimimum value of $(3x^2 - 12x + 5)^2$

thanks, mark

**skeeter** · September 2nd 2009, 05:19 AM

Originally Posted by mark

can anyone show me (in detail) how to find the mimimum value of $(3x^2 - 12x + 5)^2$

thanks, mark

note that you have an expression of the form $[f(x)]^2$

squaring anything that is non-zero results in a positive result.

the absolute possible minimum would be 0 if $f(x) = 0$ for some x value.

so ... the first thing I would do is see if $3x^2-12x+5$ equals 0 for at least one x value.

there is also a method using calculus, but since you posted this in the algebra section, I assumed you have no knowledge in that area.

**mark** · September 2nd 2009, 05:28 AM

no i have no knowledge of calculus, i haven't got that far with my studying yet. yes i tried putting x as 0 and thought that since there was a 5 left you'd have to square that and it would give 25, leaving that as the minimum value. but the answer in the book says 0 is the minimum value of that expression. i thought maybe you would have to expand the brackets or something to find the value, but really don't know what i'm doing

if anyone can show me a method of arriving at 0 using calculus or anything else can they show me please step by step how you would come to 0 with the equation $(3x^2 - 12x + 5)^2$

**RobLikesBrunch** · September 2nd 2009, 06:29 AM

I calculated the minimum value and it's greater than zero...and I remember my pre-calc teacher telling me that there is no way to calculate minimum value without calculus, and for when we saw it (on minimization problems) we simply had to use our calculators.

As far as calculating it with calculus, you need to find the first derivative, set it equal to 0 and solve for x.

As you don't have any calculus background, I'll just give you a brief overview of why this works.

A derivative is essentially a measure of how much a quantity is changing...so by what magnitude. A derivative is a function, so it describes this for every single point of your original function. A the value of the derivative at a specific point is the slope of the tangent line. A tangent line only intersects the graph at one point...so it's basically like a measure of slope (which is a measure of change) but the distance between the two points is infinitesimally small, so you get a measure of instantaneous change, rather than an average between two points.

When a function is at its lowest point, the only way it can go is up, such that the graph looks like the bottom (or top) or a parabola...because that's the parabola's lowest/highest point. You'll see that at this point, the graph looks flat..because it is. If a graph is flat and you try to take the slope of it, you get a slope of 0, because it's not changing. Same thing with the derivative..hence, we set it equal to 0 and solve.

The derivative of $(3x^2-12x+5)^2$ is $36x^3-216x^2+348x-120=(12x-24)(3x^2-12x+5)$ . We set it to 0 and solve for x:

$0=(12x-24)(3x^2-12x+5)$
Our three answers are: $x=\frac{-21^{1/2}+6}{3}$ $x=2$ and $x=\frac{21^{1/2}+6}{3}$

Note that the derivative will also give us the maximum values of an interval and the minimum values of a certain interval. However, we want the absolute minimum value...I forget what exactly they call it.

We need to plug our x-values into the original function to see which is the smallest, and it turns out that the original function was the least at $x=\frac{-21^1/2+6}{3}$ , where it equals about $10^{-26}$ ...although when I graph it I get $2.05 \cdot 10^{-23}$ which I assume is simply a result of the calculator's inability to deal with such small numbers...and in Mathematica, I get $4.43734 \cdot 10^{-31}$ ...which is probably the most accurate number.

**mark** · September 2nd 2009, 06:47 AM

lol thanks for trying to help but i think i'm gonna need to be doing this for a while longer before i understand what you just wrote. also, the answer the book gives me is that the minimum value of the expression is 0 and also in the exam we're not allowed to use a calculator.

**RobLikesBrunch** · September 2nd 2009, 06:49 AM

Originally Posted by mark

lol thanks for trying to help but i think i'm gonna need to be doing this for a while longer before i understand what you just wrote. also, the answer the book gives me is that the minimum value of the expression is 0 and also in the exam we're not allowed to use a calculator.

It's almost 0...but I'm unsure how you'd calculate it as an approximation with only algebra.

**mark** · September 2nd 2009, 06:52 AM

i guess you would have to use calculus of some sort then. i've done stuff with parabolas and graphs so maybe thats it. do you think it could have something to do with completing the square? thats what i've been doing recently...

also if it helps, the answer given for the value of x which gives the minimum value of the expression is $x = 2 \pm \sqrt {\frac{7}{3}}$

**mark** · September 2nd 2009, 07:08 AM

does anyone know how to complete the square of $(3x^2 - 12 + 5)^2$ then?

**skeeter** · September 2nd 2009, 07:25 AM

the minimum value of $(3x^2-12x+5)^2$ is 0.

$<br /> 3x^2-12x+5 = 0<br />$

completing the square is messy ... use the quadratic formula ...

$<br /> x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}<br />$

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