# Thread: Calculus Assignment - HELP!!!

1. ## 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.

2. 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.
$\displaystyle f(x)=\frac{ax+b}{(x-1)(x-4)}$

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

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

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

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

We have two small equations in 2 unknowns:

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

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

Solve for a and b

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.

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

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

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:

$\displaystyle y=mx+b$

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

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

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

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

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

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

$\displaystyle 3(x+1)^{2}(x-1)(x-3)$

3. ## ???

Does anyone else got anything for the first three questions???

4. 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.
$\displaystyle f(x) = \frac{2x^2 - 1}{x}$

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

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

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

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

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

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

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

-Dan

5. Thank you soooo much!

If I could just get an answer for number three, I will be the happiest person!

6. Originally Posted by philipsach1
3. Determine lim x-->0 (3rdroot(x-8)+2)/x
$\displaystyle \lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x}$

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

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

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

-Dan

7. ## 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.

8. 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:
$\displaystyle \sqrt[3]{0-8} + 2 = \sqrt[3]{-8} + 2 = -2+ 2 = 0$

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

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

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

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

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

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

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

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

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

-Dan

9. I'm sorry, correct me if I'm wrong but $\displaystyle (0-8)^{\frac{1}{3}}=1+\sqrt{3}i$

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

10. Hello,

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

Are you both in the same class?

EB

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

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

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

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

12. 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

13. 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.

14. 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