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 ).
If x= 9.1, then what is the value of y?
Of the original function or the derivative?
original function
y = \sqrt{ x } + 3 at x=9.1 --> 6.016620626
so we know the point (9.1, 6.016620626) lies on the function curve
P = (9,6) Q = (9.1, 6.016620626) slope of PQ = ??
that's what I don't get, I do not know what to do...
you would use the slope formula \[\LARGE m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\]
it doesn't work because both x are 9 (we cannot divide by 0)
x1 = 9 x2 = 9.1
so x2 - x1 = 9.1 - 9 = 0.1
oh i missed that one, the answer is 0.16620626 ... would that be the answer for the first one?
I entered it and it says correct!! thank you! what about b) ?
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
repeat the same steps as a) but now use x = 9.01 instead of 9.1
I meant the second part of the question
tell me what slopes you get for part a)
for every one of them of just the first one?
each one
0.16620626 -0.1666203961-0.1671322196-0.1667129887
We just do the average?
it's probably not 100% clear but what do you notice that is common to all of these slopes?
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?
sure
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 =
Do you know how to help me on this one?
|dw:1443842056603:dw|
Join our real-time social learning platform and learn together with your friends!