Critical points/ Abs. max/min values

**Oldspice1212** · May 25th 2013, 03:28 PM

Hey guys so I need help on two questions that I'm sort of stuck on.

1. Find the critical numbers of the given function: f(x) = 2x³+x²+2x

So I got f'(x) = 6x²+2x+2

Factored out the 2(3x²+x+1)

I can't factor this out anymore, so is the answer there are no critical numbers?

2. Find the absolute max and min values of f on the given interval.

√ = square root

f(t) = ³√t * (8-t), [0,8]

So I used product rule on this one and got

f'(t) = 3/2√t * (8-t) - ³√t = 0

3(8-t)-3√t = 0

So, I'm uncertain about the critical numbers here as well are they 8 and 0?

**Shakarri** · May 25th 2013, 03:41 PM

1) If you sketch a cubic equation you will see that it has no maximum or minimum.

2) There's a mistake in your differentiation of t^1/3

**Oldspice1212** · May 25th 2013, 04:46 PM

Originally Posted by Shakarri

1) If you sketch a cubic equation you will see that it has no maximum or minimum.

2) There's a mistake in your differentiation of t^1/3

Yeah I just noticed the derivative was wrong for number 2.

For 1 do I just leave my answer as it is? and write no critical points then?

**ibdutt** · May 25th 2013, 09:13 PM

For the first one the reason is that the quadratic 3x^2+x+1 does not have real roots. Its discriminant is negative so it has imaginary roots and hence no critical points.

**ibdutt** · May 25th 2013, 09:21 PM

**Oldspice1212** · May 25th 2013, 09:29 PM

Originally Posted by ibdutt

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Originally Posted by ibdutt

For the first one the reason is that the quadratic 3x^2+x+1 does not have real roots. Its discriminant is negative so it has imaginary roots and hence no critical points.

Have I told you how much I love you?

Haha, thanks a lot for both of the questions. You are the best!

**Oldspice1212** · May 25th 2013, 09:56 PM

So the only critical point for question 2 will be 0, and when looking for absolute max/ min, with the intervals, the f(x) would give 0 for both the numbers in the interval. So, is it possible to have a abs. max/min on the same point?

**x3bnm** · May 26th 2013, 01:08 PM

Originally Posted by Oldspice1212

So the only critical point for question 2 will be 0, and when looking for absolute max/ min, with the intervals, the f(x) would give 0 for both the numbers in the interval. So, is it possible to have a abs. max/min on the same point?

No same point can't be both absolute min and max at the same time.

First you find the critical points $c = \{c_1, c_2, c_3, ...\}$ by letting $\frac{d(f(x))}{dx} = 0$ and solving for $x$

then take one critical point and say it's $c$ and do second derivative test. Second derivative test is:

if $\frac{d(f(c))}{dx^2} > 0$ then $c$ is absolute/local minimum.
or on the other hand if $\frac{d(f(c))}{dx^2} < 0$ then $c$ is absolute/local maximum.

Now if this second derivative test shows that you've multiple mins and/or maxes or you're asked to find local min/max in an interval ( $x = a$ to $x = b$ ) then plugin the values of critical points and/or both ends of interval into $f(x)$ and see which one is max and which one is min.

If second derivative of a function doesn't exist then use First derivative test which you can find by using Google.

Hope this helps.

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