The point P(9 , 6 ) lies on the curve y = \sqrt{ x } + 3. Let Q be the point (x, \sqrt{ x }+ 3 ).
a.) Find the slope of the secant line PQ for the following values of x(Answers here should be correct to at least 6 places after the decimal point.)
If x= 9.1, the slope of PQ is:
If x= 9.01, the slope of PQ is:
If x= 8.9, the slope of PQ is:
If x= 8.99, the slope of PQ is:
b.) Based on the above results, estimate the slope of the tangent line to the curve at P(9 , 6 ).

- anonymous

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- jim_thompson5910

If x= 9.1, then what is the value of y?

- anonymous

Of the original function or the derivative?

- jim_thompson5910

original function

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## More answers

- anonymous

y = \sqrt{ x } + 3 at x=9.1 --> 6.016620626

- jim_thompson5910

so we know the point (9.1, 6.016620626) lies on the function curve

- jim_thompson5910

P = (9,6)
Q = (9.1, 6.016620626)
slope of PQ = ??

- anonymous

that's what I don't get, I do not know what to do...

- jim_thompson5910

you would use the slope formula
\[\LARGE m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\]

- anonymous

it doesn't work because both x are 9 (we cannot divide by 0)

- jim_thompson5910

x1 = 9
x2 = 9.1

- jim_thompson5910

so x2 - x1 = 9.1 - 9 = 0.1

- anonymous

oh i missed that one, the answer is 0.16620626 ... would that be the answer for the first one?

- anonymous

I entered it and it says correct!! thank you! what about b) ?

- jim_thompson5910

y1 = 6
y2 = 6.016620626
y2 - y1 = 6.016620626 - 6 = 0.016620626
divide that by 0.1 and you get 0.016620626/0.1 = 0.16620626
so it looks good

- jim_thompson5910

repeat the same steps as a) but now use x = 9.01 instead of 9.1

- anonymous

I meant the second part of the question

- jim_thompson5910

tell me what slopes you get for part a)

- anonymous

for every one of them of just the first one?

- jim_thompson5910

each one

- anonymous

0.16620626 -0.1666203961-0.1671322196-0.1667129887

- anonymous

We just do the average?

- jim_thompson5910

it's probably not 100% clear but what do you notice that is common to all of these slopes?

- anonymous

I tried the average of them and it is correct! Thank you so much, I appreciate it greatly! Could you help me with another quick question?

- jim_thompson5910

sure

- anonymous

Let f(x) = 3x^{4}-4.
(a) Use the limit process to find the slope of the line tangent to the graph of f at x = 3.
Slope at x = 3: (I got 324 but i dont know if it is correct)
(b) Find an equation of the line tangent to the graph of f at x = 3.
Tangent line: y =

- anonymous

Do you know how to help me on this one?

- triciaal

|dw:1443842056603:dw|

- jim_thompson5910

keep following @triciaal steps to find the value of f(3)
if x = 3, then y = ??

- anonymous

y=239

- jim_thompson5910

so we know (3,239) lies on the curve

- anonymous

and the slope of 324 was correct?

- jim_thompson5910

yes

- jim_thompson5910

so you know this line has a slope of 324 and goes through (3,239)

- anonymous

so y-y1=m(x-x1) ?

- jim_thompson5910

yes

- jim_thompson5910

m = 324
(x1,y1) = (3,239)

- anonymous

How do you rearrange it so that the equation looks like y= mx+b ??

- jim_thompson5910

you just need to solve for y and simplify

- anonymous

It worked! I just made a sign mistake at the end.... @jim_thompson5910 I appreciate your help greatly! you saved my grade! @triciaal Thank you also for helping me I appreciate it

- jim_thompson5910

you're welcome

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