# Calculus Assignment - HELP!!!

• Nov 24th 2006, 12:25 PM
philipsach1
Calculus Assignment - HELP!!!
If anyone could help me with these questions, that would be really really great.

1. Determine the derivative of f(x)=(2x^2-1)/x from first principles. Use the quotient rule to verify your answer.

2. The graph of has a f(x)=(ax+b)/((x-1)(x-4)) horizontal tangent line at (2,-1). Determine the values a and b.

3. Determine lim x-->0 (3rdroot(x-8)+2)/x

4. The tangent line to the curve defined by g(x)=(x^4-2x^3+3)/(-6x) at x=-1 intersects the curce at two other points P and Q. Determine the coordinates of P and Q, algebraically. Illustrate this situation with a graph.

• Nov 24th 2006, 01:23 PM
galactus
Quote:

2. The graph of has a f(x)=(ax+b)/((x-1)(x-4)) horizontal tangent line at (2,-1). Determine the values a and b.
$f(x)=\frac{ax+b}{(x-1)(x-4)}$

Setting f(x) =-1 and subbing in x=2, we get $f(x)=-a-\frac{b}{2}$

$f'(x)=\frac{-ax^{2}+2bx-4a-5b}{(x-1)^{2}(x-4)^{2}}$

Sub in x=2 into f'(x), we get $f'(x)=\frac{b}{4}$

Since the line is horizontal, we have $\frac{b}{4}=0$

We have two small equations in 2 unknowns:

$-a-\frac{b}{2}=-1$

$\frac{b}{4}=0$

Solve for a and b

Quote:

Originally Posted by philipsach1
4. The tangent line to the curve defined by g(x)=(x^4-2x^3+3)/(-6x) at x=-1 intersects the curce at two other points P and Q. Determine the coordinates of P and Q, algebraically. Illustrate this situation with a graph.

$f(x)=\frac{x^{4}-2x^{3}+3}{-6x}$

$f'(x)=\frac{-3x^{4}+4x^{3}+3}{6x^{2}}$

Sub x=-1 into f(x) and we find y=1

Sub x=-1 into f'(x) to find the slope at x=-1. We get m=-2/3

Use the given coordinates and the new found slope to find the equation of our line:

$y=mx+b$

$1=(\frac{-2}{3})(-1)+b$

$b=\frac{1}{3}$

Therefore, the equation of the line is $y=\frac{-2}{3}x+\frac{1}{3}$

Set f(x) and the line equation equal and solve for x:

$\frac{x^{4}-2x^{3}+3}{-6x}=\frac{-2}{3}x+\frac{1}{3}$

$3x^{4}-6x^{3}-12x^{2}+6x+9=0$

$3(x+1)^{2}(x-1)(x-3)$
• Nov 25th 2006, 11:05 AM
philipsach1
???
Does anyone else got anything for the first three questions???
• Nov 25th 2006, 11:19 AM
topsquark
Quote:

Originally Posted by philipsach1
1. Determine the derivative of f(x)=(2x^2-1)/x from first principles. Use the quotient rule to verify your answer.

$f(x) = \frac{2x^2 - 1}{x}$

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$f'(x) = \lim_{h \to 0} \frac{ \frac{2(x+h)^2 - 1}{x + h} - \frac{2x^2 - 1}{x} }{h}$

$f'(x) = \lim_{h \to 0} \frac{(2(x+h)^2 - 1)x - (2x^2 - 1)(x + h)}{hx(x+h)}$

$f'(x) = \lim_{h \to 0} \frac{2x^3 + 4x^2h + 2xh^2 - x - 2x^3 - 2x^2h + x + h}{hx(x+h)}$

$f'(x) = \lim_{h \to 0} \frac{2x^2h + 2xh^2 + h}{hx(x+h)}$

$f'(x) = \lim_{h \to 0} \frac{2x^2 + 2xh + 1}{x(x+h)}$

$f'(x) = \frac{2x^2 + 1}{x^2}$

-Dan
• Nov 25th 2006, 11:27 AM
philipsach1
Thank you soooo much!

If I could just get an answer for number three, I will be the happiest person!
• Nov 25th 2006, 11:28 AM
topsquark
Quote:

Originally Posted by philipsach1
3. Determine lim x-->0 (3rdroot(x-8)+2)/x

$\lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x}$

This is in the form of $\frac{0}{0}$ so use L'Hopital's rule:

$\lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x} = \lim_{x \to 0} \frac{ \frac{1}{3} \cdot \frac{1}{\sqrt[3]{(x-8)^2}} }{1}$

= $\frac{1}{3} \cdot \frac{1}{\sqrt[3]{(-8)^2}} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$

-Dan
• Nov 26th 2006, 08:09 PM
philipsach1
Sorry
Sorry again, but could you explain to me how you used the l'Hopital rule in question 3? i can't figure out the process.
• Nov 27th 2006, 04:06 AM
topsquark
Quote:

Originally Posted by philipsach1
Sorry again, but could you explain to me how you used the l'Hopital rule in question 3? i can't figure out the process.

No problem!

Note that, for x = 0 the numerator becomes:
$\sqrt[3]{0-8} + 2 = \sqrt[3]{-8} + 2 = -2+ 2 = 0$

So for x = 0 the fraction takes on the form $\frac{0}{0}$. This is one of the conditions under which we may use L'Hopital's rule.

L'Hopital's rule says that:
$\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}$ when $\lim_{x \to a}f(x) = 0$ and $\lim_{x \to a}g(x) = 0$.

So:
$\lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x} \to \frac{0}{0}$

We need to take the derivative of both the numerator and denominator.

Numerator:
$\frac{d}{dx}(\sqrt[3]{x-8} + 2 ) = \frac{d}{dx}((x - 8)^{1/3} + 2)$

= $\frac{1}{3}(x - 8)^{-2/3} = \frac{1}{3\sqrt[3]{(x - 8)^2}}$

Denominator:
$\frac{d}{dx}x = 1$

So:
$\lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x} = \lim_{x \to 0} \frac{ \frac{1}{3} \cdot \frac{1}{\sqrt[3]{(x-8)^2}} }{1}$

= $\frac{1}{3} \cdot \frac{1}{\sqrt[3]{(-8)^2}} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$

-Dan
• Nov 27th 2006, 04:26 AM
galactus
I'm sorry, correct me if I'm wrong but $(0-8)^{\frac{1}{3}}=1+\sqrt{3}i$

I ran this through the calc to be sure, but I get an undefined limit.
• Nov 27th 2006, 04:57 AM
earboth
Hello,

have a look here: http://www.mathhelpforum.com/math-he...jmailloux.html

Are you both in the same class?

EB
• Nov 27th 2006, 05:03 AM
earboth
Quote:

Originally Posted by galactus
I'm sorry, correct me if I'm wrong but $(0-8)^{\frac{1}{3}}=1+\sqrt{3}i$

I ran this through the calc to be sure, but I get an undefined limit.

Hello, galactus,

$(0-8)^{\frac{1}{3}}=\sqrt[3]{-8}=\sqrt[3]{(-2)^3}=-2$

Quote:

Originally Posted by galactus

I ran this through the calc to be sure, but I get an undefined limit.

The limit should be calculated with x approaching zero.

EB
• Nov 27th 2006, 05:16 AM
topsquark
Quote:

Originally Posted by galactus
I ran this through the calc to be sure, but I get an undefined limit.

I'm curious. The screenshot looks like a TI-92, but mine will calculate the limit. What calculator are you using?

-Dan
• Nov 27th 2006, 05:29 AM
galactus
I also have a 92. I used x approaching 0. It looks funny in the post.

My apologies. I had the complex format on my calculator set to rectangular instead of real. That made the difference.
• Nov 27th 2006, 06:04 AM
topsquark
Quote:

Originally Posted by galactus
I also have a 92. I used x approaching 0. It looks funny in the post.

My apologies. I had the complex format on my calculator set to rectangular instead of real. That made the difference.

No apologies necessary. :)

-Dan