Thread: Critical points/ Abs. max/min values

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?

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

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?

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.

Originally Posted by ibdutt

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!

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?

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 $\displaystyle c = \{c_1, c_2, c_3, ...\}$ by letting $\displaystyle \frac{d(f(x))}{dx} = 0$ and solving for $\displaystyle x$

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

if $\displaystyle \frac{d(f(c))}{dx^2} > 0$ then $\displaystyle c$ is absolute/local minimum.
or on the other hand if $\displaystyle \frac{d(f(c))}{dx^2} < 0$ then $\displaystyle c$ 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 ( $\displaystyle x = a$ to $\displaystyle x = b$ ) then plugin the values of critical points and/or both ends of interval into $\displaystyle f(x)$ 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.