On which two intervals is the function increasing

Question

For x E [-11,11] the function f is defined by 

f(x)=x⁶(x-2)⁵

1) On which two intervals is the function increasing?

2)Find the region in which the function is positive

3)Where does the function achieve its minimum?

Answer 1

Extrema (maximum and minimum) come from the first derivative of an equation. For this equation, finding the derivative requires either multiplying the polynomial out (ICK) or using the Product Rule. I'll opt for the latter.

 

f(x) = x⁶ (x-2)⁵

 

f'(x) = 6x⁵ (x-2)⁵ + x⁶ * 5(x-2)⁴

 

You'll need the zeros of this derivative, so it's time to factor...

 

f'(x)  = 6x⁵ (x-2)⁵ + 5x⁶(x-2)⁴

f'(x) = x⁵(x-2)⁴ [6(x-2) + 5x]

f'(x) = x⁵(x-2)⁴(11x - 12)

 

So our zeros are x = 0, x = 2, and x = 12/11. These are the possible places in the graph where it changes direction, which is the same as changing sign from negative to positive or vice versa. We'll use the first derivative test to find out if areas have a positive or negative slope. Personally, I've always taught this by using a number line, as it's a quick way to visualize and summarize the information.

 

f'(x)   <-----0---------12/11---------2---------->

 

Once you have your line set up, you just need to test a number within each interval by putting it in the derivative and computing its sign (negative or positive). Note that you DON'T have to compute the value of the derivative; only its sign. So let's take a look.

 

There are three parts to this derivative: x⁵, (x-2)⁴, and (12x - 11). We can look at each of those terms and determine the sign at any given point without actually computing the value. Say we test the interval below 0...so let's test x = -1. Don't bother computing the value; just determine the sign.

 

x⁵ = negative (in this case, -1)

(x - 2)⁴ = positive (anything to the 4th power is positive, so this is positive regardless of x)

(11x - 12) = negative (you'd have a negative - a negative, which gives you a bigger negative)

 

So when x = -1, we'd have f'(x) = negative * positive * negative... which will result in a positive number. This means....

 

            pos(+)

f'(x)   <--------0-------11/12-------2------>

 

Now let's check the next interval, between 0 and 11/12. For thought's sake, 1/2 lies within that interval.

x⁵ = pos (in this case, 1)

(x-2)⁴ = pos (established above, always positive)

(12x - 11) = neg (in this case, -5)

 

So we have f'(x) = pos * pos * neg = neg

 

Updating...

 

             pos(+)    neg(-)

f'(x)   <---------0---------11/12--------2------->

 

Testing the interval between 11/12 and 2 (say, with x = 1) will give you pos*pos*pos = pos.  Testing the interval over 2 (say, with 3) will give you pos* pos*pos = pos. So our number line summarizing the first derivative...

 

            pos(+)       neg(-)           pos(+)      pos(+)
f'(x)   <---------0---------11/12----------2------->

 

We are limited by the interval they provided, so rather than an infinite number line, it stops...

 

               pos(+)    neg(-)           pos(+)    pos(+)
f'x)    -11---------0---------11/12----------2---------11

 

Alright! This answers the first question, except there is one glitch. The function is increasing on the interval [-11,0) and (11/12,11]. However, this is actually incorrect, as the function is NOT increasing at x = 2... at that point, the slope of the function is zero. So technically there are THREE intervals, not two...

[-11,0), (11/12, 2), and (2, 11].

 

Note that the -11 and 11 have square brackets because those numbers can be included in the interval. The numbers that came from the derivative cannot, since at those points the graph is neither increasing or decreasing. That's why they have the rounded parentheses.

 

Part 2... for which region is the function positive?

 

This part isn't asking about the derivative. It's asking about the original function -- where is it positive or negative? The original function has zeros at x = 0 and x = 2. I set up a number line the same way....

 

f(x)  -11------0-------2------11

 

... and test the same way, this time in the original function. Note that x⁶ will always be positive, so we only have to worry about (x-2)⁵. This guy is negative when x < 2 and positive when x > 2. So our number line looks like this:

 

           neg     neg     pos

f(x)  -11------0-------2------11

 

This means the function is positive on the interval (2,11].

 

Part 3....

 

A minimum occurs when the graph changes from going downhill (slope = derivative = negative) to going uphill (derivative = positive). Looking at our derivative number line...

 

             pos(+)    neg(-)          pos(+)      pos(+)
f'x) -11---------0---------11/12----------2---------11

 

So the only place that change occurs is at x = 11/12. To provide a point, we have to stick 11/12 in the original equation for x...

 

f(11/12) = (11/12)⁶(11/12-2)⁵

f(11/12) = .59329 * -1.49214

f(11/12) = -.88527

 

So we have the point (11/12, -.88527).

 

Now, normally this would be a relative minimum. HOWEVER, since they gave you an interval, one of those endpoints COULD have a lower y value. So we need to check those to see if their y value is below -.88527.

 

f(-11) = very close to zero (-.0000000000065776)

f(11) = very high (~ 104608905500)

 

Comparing these y values, x = 11/12 does in fact yield the lowest y value. So that point is the minimum on this interval.

 

Woot...that was long! :)