# Thread: Slope of tangent #1

1. ## Slope of tangent #1

i) Find the points on the graph of y=(1/3)x^3 -5x-(4/x) at which the slope of the tangent is horizontal.

ii) find the slope of the demand curve D(p)=20/((sqrt p -1)), p>1, at point (5,10).

I need help with the two questions above, I'm doing practice questions with the slope of tangent but these last two questions I'm stuck on. Please help me or guide me in the right direction so I can finish these two. Thanks for your help in advance!

2. Originally Posted by skeske1234
i) Find the points on the graph of y=(1/3)x^3 -5x-(4/x) at which the slope of the tangent is horizontal.

ii) find the slope of the demand curve D(p)=20/((sqrt p -1)), p>1, at point (5,10).

I need help with the two questions above, I'm doing practice questions with the slope of tangent but these last two questions I'm stuck on. Please help me or guide me in the right direction so I can finish these two. Thanks for your help in advance!
1. Take the derivative. Set it to 0. Solve for x. Find the cordinates.

2. Take the derivative. Evaluate the derivative at x=5.

Remember: Derivative=Slope=rate of change

3. Ok, for the first question when you say : "1. Take the derivative. Set it to 0. Solve for x. Find the cordinates."

I am stuck with the first part. When I take the derivative, what do I put in for "a" in the first principle of derivatives formula? ie. [f(a+h) - f(a)]/h

and then after setting it to 0 and solving for x to find the coordinates.. not sure how to do that part. could you please show me how to begin it and explain more . please and thank you! very much

4. Originally Posted by skeske1234
Ok, for the first question when you say : "1. Take the derivative. Set it to 0. Solve for x. Find the cordinates."

I am stuck with the first part. When I take the derivative, what do I put in for "a" in the first principle of derivatives formula? ie. [f(a+h) - f(a)]/h

and then after setting it to 0 and solving for x to find the coordinates.. not sure how to do that part. could you please show me how to begin it and explain more . please and thank you! very much
Do you have to use the limit definition of a derivative (first principle of derivatives) or can you use differentiation rules, in this case the power rule?

5. Originally Posted by McScruffy
Do you have to use the limit definition of a derivative (first principle of derivatives) or can you use differentiation rules, in this case the power rule?

i have to use the first principle of derivatives, so how would i go about doing that for the first question?

6. Originally Posted by skeske1234
i have to use the first principle of derivatives, so how would i go about doing that for the first question?
$\displaystyle \frac{d}{dx}\left[\frac{1}{3}x^3-5x-4x^{-1}\right]=(3)\frac{1}{3}x^{3-1}-(1)5x^{1-1}-(-1)4x^{-1-1}=x^2-5+4x^{-2}$

By the power rule.

If it's first priciples you want do this

$\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

can you proceed? The answer will be the same as above.

Ah heck, here ya go. I'll stop bein' lazy.

$\displaystyle \lim_{h\to0}\frac{\overbrace{\left[\frac{1}{3}(x+h)^3-5(x+h)-\frac{4}{(x+h)}\right]}^{f(x+h)}-\overbrace{\left[\frac{1}{3}x^3-5x-\frac{4}{x}\right]}^{f(x)}}{h}$

7. Originally Posted by skeske1234
i have to use the first principle of derivatives, so how would i go about doing that for the first question?
$\displaystyle \lim_{h\rightarrow0}\frac{\left(\frac{1}{3}(x+h)^3-5(x+h)-\frac{4}{x+h}\right)-\left(\frac{1}{3}x^3-5x-\frac{4}{x}\right)}{h}$

You beat me that time VonNemo19!

You would in turn work out the derivative, set that equal to zero and solve for x. That is the x-value of the point at which the tangent line is horizontal.

8. Originally Posted by VonNemo19
$\displaystyle \frac{d}{dx}\left[\frac{1}{3}x^3-5x-4x^{-1}\right]=(3)\frac{1}{3}x^{3-1}-(1)5x^{1-1}-(-1)4x^{-1-1}=x^2-5+4x^{-2}$

By the power rule.

If it's first priciples you want do this

$\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

can you proceed? The answer will be the same as above.

Ah heck, here ya go. I'll stop bein' lazy.

$\displaystyle \lim_{h\to0}\frac{\overbrace{\left[\frac{1}{3}(x+h)^3-5(x+h)-\frac{4}{(x+h)}\right]}^{f(x+h)}-\overbrace{\left[\frac{1}{3}x^3-5x-\frac{4}{x}\right]}^{f(x)}}{h}$

once i set it to 0 and solve for x, how do i find the coordinates? I have x^4-15x^2-12=0

9. Originally Posted by skeske1234
once i set it to 0 and solve for x, how do i find the coordinates? I have x^4-15x^2-12=0
Plug that x-value back into the original equation to get your y-value. That will give you the (x, y) coordinates of the point where the tangent line is horizontal.

10. Originally Posted by McScruffy
Plug that x-value back into the original equation to get your y-value.
Indeed, Mcscruffy! Indeed!

What he said!

11. Originally Posted by VonNemo19
Indeed, Mcscruffy! Indeed!

What he said!

thanks for your help guys but.. i am still stuck.. i can't seem to find the derivative when I use the [f(x+h)-f(x) ]/h formula.. i cant seem to cancel out the h in the denominator.. could you please show me how to do this? thanks

12. Originally Posted by skeske1234
thanks for your help guys but.. i am still stuck.. i can't seem to find the derivative when I use the [f(x+h)-f(x) ]/h formula.. i cant seem to cancel out the h in the denominator.. could you please show me how to do this? thanks
Check your algebra one more time, my friend.

I'll do something similar to show you.

$\displaystyle \lim_{h\to0}\frac{\left[(x+h)^2+\frac{4}{x+h}\right]-\left(x^2+\frac{4}{x}\right)}{h}$

$\displaystyle =\lim_{h\to0}\frac{\left[x^2+2hx+h^2+\frac{4}{x+h}\right]-\left(x^2+\frac{4}{x}\right)}{h}$

$\displaystyle =\lim_{h\to0}\frac{2hx+h^2+\frac{4}{x+h}-\frac{4}{x}}{h}$

$\displaystyle =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{\ frac{4}{x+h}-\frac{4}{x}}{h}$

$\displaystyle =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{\ frac{4x-4(x+h)}{x(x+h)}}{h}$

$\displaystyle =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{4 x-4x-4h}{h(x)(x+h)}$

Can you take it from here? again... this is not the answer to your problem. It's ver similar though.

13. Originally Posted by skeske1234
thanks for your help guys but.. i am still stuck.. i can't seem to find the derivative when I use the [f(x+h)-f(x) ]/h formula.. i cant seem to cancel out the h in the denominator.. could you please show me how to do this? thanks
Long story short, I wound up with:

$\displaystyle ...$ (obviously there are a couple of steps before this)

$\displaystyle \lim_{h\to0}\frac{x^4h-5x^2h-5xh^2+4h}{x^2h+xh^2}$

$\displaystyle \lim_{h\to0}\frac{x^4-5x^2-5xh+4}{x^2+xh}$

which is: $\displaystyle x^2-5+\frac{4}{x^2}$, which is the derivative.

14. Originally Posted by VonNemo19
Check your algebra one more time, my friend.

I'll do something similar to show you.

$\displaystyle \lim_{h\to0}\frac{\left[(x+h)^2+\frac{4}{x+h}\right]-\left(x^2+\frac{4}{x}\right)}{h}$

$\displaystyle =\lim_{h\to0}\frac{\left[x^2+2hx+h^2+\frac{4}{x+h}\right]-\left(x^2+\frac{4}{x}\right)}{h}$

$\displaystyle =\lim_{h\to0}\frac{2hx+h^2+\frac{4}{x+h}-\frac{4}{x}}{h}$

$\displaystyle =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{\ frac{4}{x+h}-\frac{4}{x}}{h}$

$\displaystyle =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{\ frac{4x-4(x+h)}{x(x+h)}}{h}$

$\displaystyle =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{4 x-4x-4h}{h(x)(x+h)}$

Can you take it from here? again... this is not the answer to your problem. It's ver similar though.
ok, about the fourth line down on yours.. why did you split up the question to two limh>0 + limh>0?
sorry, i still can't seem to get the h to cancel out on mine.. I will show you what I have

15. Originally Posted by skeske1234
ok, about the fourth line down on yours.. why did you split up the question to two limh>0 + limh>0?
sorry, i still can't seem to get the h to cancel out on mine.. I will show you what I have
A strategic move that made my life easier. It is valid because of the theorems on limits says it is.

I first rewrote the quotient as two separate ones by the basic division identity

$\displaystyle \frac{a+b+c+d}{h}=\frac{(a+b)+(c+d)}{h}=\frac{a+b} {h}+\frac{c+d}{h}$

And since the limit of a sum is the sum of the limits, I could "distribute" the limit to both quotients. (I probably shouldn't use the word distribute, but who cares)

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