# Thread: Extremes of Given Functons

1. ## Extremes of Given Functons

Need help on the following I can do easy ones of finding extremes like -x^3+2x^2 but these problems I'm having trouble figuring out what to do. Also how do I find absolute values?

a. Find the Interval on which the function is increasing and deceasing.
b. Then identify the functions local extreme values if any.
c. Which, if any, are absolute values?

1. x sq.(8-x^2)
2. x^(1/3) (x+8)
3. e^2x + e^-x

2. Originally Posted by goldenroll
Need help on the following I can do easy ones of finding extremes like -x^3+2x^2 but these problems I'm having trouble figuring out what to do. Also how do I find absolute values?

a. Find the Interval on which the function is increasing and deceasing.
b. Then identify the functions local extreme values if any.
c. Which, if any, are absolute values?

1. x sq.(8-x^2)
2. x^(1/3) (x+8)
3. e^2x + e^-x
The first step is to calculate the derivative of each function. Can you do this? If not, where do you get stuck?

3. Originally Posted by goldenroll
Need help on the following I can do easy ones of finding extremes like -x^3+2x^2 but these problems I'm having trouble figuring out what to do. Also how do I find absolute values?

a. Find the Interval on which the function is increasing and deceasing.
b. Then identify the functions local extreme values if any.
c. Which, if any, are absolute values?

1. x sq.(8-x^2)
2. x^(1/3) (x+8)
3. e^2x + e^-x
If you can do easy ones, you can do "hard" ones. The idea is the same:

Take the first derivative and set it to 0 to find the critical values of the function.

Choose a test value in each of the intervals partitioned by the critical values.

Determine where f '(x) is increasing or decreasing by evaluating f ' at the test values. If f '>0, then f is inc. If f '<0 then f is dec.

I'll give you some hints for your problems:

For 1. Differentiate using the product rule and chain rule.
For 2. product rule
For 3. Hint: $\frac{d}{dx}[e^u]=e^u\cdot{u'}$. In your case, u is a function of x.

4. well I think its more that I'm just getting confused (summer school cal just feels like huge load of information and hard to remember the rules) but here is what I gotten for my first question.

x/2(8-x^2) right and then I times that by 2(8-x^2)/2(8-x^2) which is part of the chain rule (thinking this is were I'm messing up, then 2x(8-x^2)/2(8-x^2) then the locals are x=0, plus or minus sq. 8?

2.
for the second question I did this 1/3 x^(-2/3) (x+8) + x^(1/3)1 = 4x+8/3x^(2/3) = 4(x+2)/3x^(2/3) = x= 0,-2

a. Incr. (-2,0), (0,inf.), Decr. on (-inf.,-2)
b. min at -2
c. Still don't quiet understand how to do ab.

5. Originally Posted by goldenroll
well I think its more that I'm just getting confused (summer school cal just feels like huge load of information and hard to remember the rules) but here is what I gotten for my first question.

x/2(8-x^2) right and then I times that by 2(8-x^2)/2(8-x^2) which is part of the chain rule (thinking this is were I'm messing up, then 2x(8-x^2)/2(8-x^2) then the locals are x=0, plus or minus sq. 8?

2.
for the second question I did this 1/3 x^(-2/3) (x+8) + x^(1/3)1 = 4x+8/3x^(2/3) = 4(x+2)/3x^(2/3) = x= 0,-2

a. Incr. (-2,0), (0,inf.), Decr. on (-inf.,-2)
b. min at -2
c. Still don't quiet understand how to do ab.
I'll walk you through it...
$\frac{d}{dx}\left[x^{1/3}(x+8)\right]=x^{1/3}\cdot1+\frac{1}{3}x^{-2/3}(x+8)]=x^{1/3}+\frac{1}{3}x^{-2/3}(x+8)$

Now we factor out the lowest power of the expont. And its usually a good idea to factor out constant fractions as well. It works to your advantage.

$=\frac{1}{3}x^{-2/3}\left[3x+(x+8)\right]=\frac{1}{3}x^{-2/3}(4x+8)$

Factor out a four

$=\frac{4}{3}x^{-2/3}(x+2)$

Now the three is in the denomiator, and you know that negative exponents will go in the denominator and become positive. Therefore

$\frac{d}{dx}\left[x^{1/3}(x+8)\right]=\frac{4(x+2)}{3\sqrt[3]{x^2}}=0\Rightarrow{\text{ critical number at }}{x}=-2$

Note: It is a good idea to use 0 as a critical number because f prime of x is undefined there AND its in the domain of f. If it were not in the domain of f, we would not cosinder it as a critical number.

So you did well. Not so hard after all...

6. Originally Posted by goldenroll
well I think its more that I'm just getting confused (summer school cal just feels like huge load of information and hard to remember the rules) but here is what I gotten for my first question.

x/2(8-x^2) right and then I times that by 2(8-x^2)/2(8-x^2) which is part of the chain rule (thinking this is were I'm messing up, then 2x(8-x^2)/2(8-x^2) then the locals are x=0, plus or minus sq. 8?

[snip]
I will give a detailed explanation of differentiating $y = x \sqrt{8 - x^2}$:

Use the product rule: $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

$u = x \Rightarrow \frac{du}{dx} = 1$

$v = \sqrt{8 - x^2} \Rightarrow \frac{dv}{dx} = .... ?$

-----------------------------------------------------------------------------

Treat finding the derivative of $v = \sqrt{8 - x^2}$ as a brand new problem.

Use the chain rule: $\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx}$.

Let $w = 8 - x^2$. Then:

1. $\frac{dw}{dx} = -2x$, and

2. $v = \sqrt{w} = w^{1/2} \Rightarrow \frac{dv}{dw} = \frac{1}{2} w^{-1/2} = \frac{1}{2 w^{1/2}} = \frac{1}{2 \sqrt{w}} = \frac{1}{2 \sqrt{8 - x^2}}$.

Substitute into the chain rule formula:

$\frac{dv}{dx} = \frac{-2x}{2 \sqrt{8 - x^2}} = \frac{-x}{\sqrt{8 - x^2}}$.

Recruite this result into the main working:

-----------------------------------------------------------------------------

$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

$u = x \Rightarrow \frac{du}{dx} = 1$

$v = \sqrt{8 - x^2} \Rightarrow {\color{red}\frac{dv}{dx} = \frac{-x}{\sqrt{8 - x^2}} }$

Therefore:

$\frac{dy}{dx} = \frac{-x^2}{\sqrt{8 - x^2}} + \sqrt{8 - x^2}$

$= \sqrt{8 - x^2} - \frac{x^2}{\sqrt{8 - x^2}}$

$= \frac{8 - x^2}{\sqrt{8 - x^2}} - \frac{x^2}{\sqrt{8 - x^2}}$

$= \frac{8 - 2x^2}{\sqrt{8 - x^2}}$.

Note that every little step needs to be done if you expect to get the correct answer. No short cuts.