Please read...CALCULUS...?

Question

Hi...
I've got some calculus questions that are really confusing me...I'm having to teach myself, and I'm not the best at it...So please help...

1. Find the largest open interval where the function is decreasing.
f(x)=square root of (4-x)
I got the answer decreasing at (4,infinity)

2. Identify the function's extreme values in the given domain and
say where they are assumed. Tell which of the values, if any, are absolute.
x^2 -4x, (-infinity is less than or = to x less than or = to 4)
Iot the answer local min. at (2,-4) local and absolute max. at (4,0)
I'm sure I got this one wrong...

3. Determine the location of each local extremum for the function.
x^3 +5x^2-13x+1
( got local max at 1, and local min. at 1)

4. Using the derivative shown, determine the intervals where the function is increasing or decreasing.
(1-x)(5-x)
(I got decreasing at (-infinity, 1) and increasing on (5, infinity))

So please help if you can...I just need to figure this stuff out..THANKS

Answer 1

1)
To find increasing and decreasing of a function, you need to take the derivative:

sqrt(4-x) can be re-written as (4-x)^(1/2)

Taking the power rule and chain rule, you get (1/2)(4-x)^(-1/2)(-1)

Simplifying, you get -1/(sqrt(4-x)).

To determine increasing or decreasing of a function, look at the derivative. If the derivative is positive, the function is increasing and if the derivative if negative, the function is decreasing. Looking from the derivative, you can conclude that the derivative is only defined for x < 4 because otherwise, you would have a negative square root which has no real value. Thus, for all x < 4, you can conclude that the derivative is negative (try plugging in a value for x).

Therefore, the function is decreasing from (-infinity, 4)
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2)
Take the derivative of the function:

f(x) = x^2 - 4x
f'(x) = 2x - 4

Setting the derivative equal to zero locates your extreme values:

2x - 4 = 0 ---------------> x = 2

Now perform the number line test to see how the function behaves to the right and left of x = 2. If you plug in a value to the left of 2, let's say x = 1, you get a negative derivative, thus the function is decreasing. If you plug in a values to the right of 2, let's say x = 3, the derivative is positive. Therefore, you can conclude from these findings that x = 2 is a minimum. Solving for the actual point, plug in x = 2 into original function. You get y = -4.

Since the function extends to infinity in the y-direction for negative and positive infinty of x, (2, - 4) is an ABSOLUTE MINIMUM
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3)
Again, take the derivative:

f(x) = x^3 + 5x^2 -13x + 1
f'(x) = 3x^2 + 10x - 13

Now set the derivative equal to zero to find extreme values:

3x^2 + 10x - 13 = 0
(x - 1)(3x + 13) = 0

Therefore, you have extreme points at x = 1, -13/3

Now do number line test like above to see if you get max or min.

Remember,
negative->positive change = minimum
positive->negative change = maximum

x = 1
to the left (like x = 0) ---------> negative
to the right (like x = 2) -------> positive

Thus, at x = 1, you have a minimum. The exact point is (1,-6)

x = -13/3
to the left (like x = -5) ---------> positive
to the right (like x = -4) -------> negative

Thus, at x = -13/3, you have a maximum. Solve for y to get the exact point.

Now determine values at x = -+ infinity

At x = +infinity, y = +infinity
At x = -infinity, y = -infinity.

Therefore, since at x = 1 and x= -13/3 give you y values that are not the extreme values, these are local extremes. Thus, x = 1 is a local minimum and x = -13/3 is a local maximum
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4)

If you plug in x = 0, you get a positive value thus it is increasing from -infinity to 1. if you plug in x = 6, you get a positive value, thus the function is increasing from 5 to infinity. If you plug in a value between 1 and 5 (like 3), you get a negative value, thus the function is decreasing from x = (1,5)
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Hope this helps

Answer 2

1) Well, 4-x has to be > than 0 or you get into imaginary numbers, so if 4-x>0 then x<4, so the interval where the function is decreasing would be (- infinity, 4)