HELP WITH calculus!?

Question

Consider the function f(x) = −2 x3 + 30 x2 − 96 x + 1. For this function there are three important intervals: (−∞, A], [A,B], and [B,∞) where A and B are the critical numbers.
Find A
and B
For each of the following intervals, tell whether f(x) is increasing (type in INC) or decreasing (type in DEC).
(−∞, A]:
[A,B]:
[B,∞):

Answer 1

f(x) = −2 x^3 + 30 x^2 − 96 x + 1

Use the first derivative to determine whether f'(x) is increasing or decreasing since f'(x) really represents the slope at every point on the line. i.e. If f'(x) is negative, the slope is negative, and the function would be decreasing.

Let's start by finding the derivative:

f'(x) = −6 x^2 + 60x − 96

The critical numbers (and also your local minima and maxima) occur when the slopes change from either negative to positive or positive to negative. Realize that this occurs when the slope equals zero.

Hence:

f'(x) = 0 = −6 x^2 + 60x − 96
0 = x^2 - 10x + 16
0 = (x - 2)(x - 8)
x = 2, x = 8

Since 2 and 8 are the critical values, A = 2 and B = 8

Therefore, we need to check if the slope is negative or positive for the intervals given in the question.

(−∞, 2], [2,8], and [8,∞)

We do this by choosing values besides the critical numbers that are contained within each interval:

(−∞, 2]
Choose x = 0:
f'(0) =−6 (0)^2 + 60(0) − 96 = -96
f'(x) is negative, and f(x) is therefore decreasing

[2,8]
Choose x = 6:
f'(0) =−6 (6)^2 + 60(6) − 96 = 480
f'(x) is positive, and f(x) is therefore increasing

[8,∞)
Choose x = 10:
f'(10) =−6 (10)^2 + 60(10) − 96 = -96
f'(x) is negative, and f(x) is therefore decreasing

In order to test for concavity, you need the second derivative. If f''(x) is negative, f(x) is concave down and if f''(x) is positive, f(x) is concave up.

Let's solve for the second derivative:


f'(x) = −6 x^2 + 60x − 96
f''(x) = −12x + 60

Once more, we need to find the critical values:

f''(x) = 0 = -12x + 60
-60 = -12x
5 = x

There is only one critical value, so we only have two intervals (−∞, 5] and [5,∞) with which to work.

(−∞, 5]
Choose x = 0:
f''(0) = -12(0) + 60 = 60
f''(x) is positive, and f(x) is therefore concave up

[5,∞)
Choose x = 10:
f''(10) = -12(10) + 60 = -60
f''(0) = -12(0) + 60 = 60
f''(x) is negative, and f(x) is therefore concave down

Answer 2

f(x) = −2 x^3 + 30 x^2 − 96 x + 1
The 1st derivative f'(x) = −6 x^2 + 60 x − 96

For f(x) to be increasing, f'(x) > 0
For f(x) to be decreasing, f'(x) < 0

Interval for increasing f(x),
−6 x^2 + 60 x − 96 > 0
6 x^2 - 60 x + 96 < 0
x^2 - 10 x + 16 < 0
(x - 2)(x - 8) < 0

So one < 0 but other > 0.

1) (x - 2) > 0 but (x - 8) < 0,
so 2 2) (x - 8) > 0 but (x - 2) < 0,
impossible as x cannot be > 8 but less than 2 at the same time.

Interval for decreasing f(x),
Similarly as shown above
(x - 2)(x - 8) > 0

So either both (x - 2) and (x - 8) < 0, or both > 0.
1) (x - 2) > 0 and (x - 8) > 0, so x > 8
2) (x - 2) < 0 and (x - 8) < 0, so x < 2

Hence the function is increasing for the interval (2,8) and it is decreasing at (−∞,2) and (8,∞).

Thus, A = 2 and B = 8.

Note:
The point x = 5 is the point of inflexion, neither maxima or minima. Here the slope changes the sign.

Answer 3

First you need to take the derivative of f(x) so you get f'(x)=-6x2+60x-96. Factoring f'(x) and setting f'(x)=0 you get -6(x-2)(x-8)=0. So you get the intervals (-∞,2], [2,8], and [8,∞) as your three intervals where A=2, B=8 are you critical numbers.
Then you can set up a table testing whether the factors, -6, (x-2), (x-8), of f'(x) give you a positive or negative value within those three intervals (leave out the critical points when you're testing so the intervals will be (-∞,2), (2,8) and (8,∞))
For the interval (-∞,2), -6 will always be negative, (x-2) is always negative, (x-8) is always negative. So multiplying the signs you get a neg x neg x neg = negative. For the interval (2,8), -6 is always negative, (x-2) will be positive, (x-8) will be negative, so neg x pos x neg = positive. For interval (8,∞),-6 is always negative, (x-2) is always positive, and (x-8) is always positive so neg x pos x pos =negative. So this tells you that f(x) is decreasing on (-∞, 2), increasing on (2,8) and decreasing on (8,∞).

A table would help you organize the above. Hope that helps.

Answer 4

i could say structure is area of engineering which signifies that a superior draw close on math is mandatory, pre-cal is is like algebra to the subsequent factor with the Pi chart and understanding about radians and attitude measures (which seems important) yet Calculus is the study of derivatives, (the slope of a line at a particular factor) form of unnecessary no count number number what field of workd your going to, in case you question me.