<p>can someone outline to me how to This problem...</p>
<p>i cant understand how to do it...</p>
<p>Find the derivative of the function and use the pt.-slope form of a line to find the equation of the tangent. Then solve both equations to find the x coordinate. Afterwards, set up the final problem by drawing a diagram.</p>
<p>also the last question on this page... or im screwed for the test</p>
<p><a href="http://www.math.neu.edu/%7Eolson/AP/L29%5B/url%5D">http://www.math.neu.edu/~olson/AP/L29</a></p>
<p>Sorry, have not gotten that far yet.</p>
<p>well i got the derivative to be 1 - (1/250x) and know point slope formula is y-y1= m (x-x1) ... but u dont know the x or y coordinates of point Q... where do i go from here</p>
<p>come on, i know theres at least 1000 people on here who have taken multi-variable calculus at age 14...this is calc I basics, im just terrrrbile</p>
<p>I just briefly glanced at it but it should be as follows.
Just take the derivative of the curve, that equals the slope m. Then use the given points on the line to create the equation of the line using point slope. Then you want to set the tangent line equal to the curve and solve for x. That should be your answer (a). (b) is just the equation of the tangent you already found.</p>
<p>what about the last question on this page...<a href="http://www.math.neu.edu/%7Eolson/AP/L29%5B/url%5D">http://www.math.neu.edu/~olson/AP/L29</a>
its extremely difficult</p>
<p>For part A:
I think the equation is 2-3x=f(x). The y-intercept is 2, because that is the value at f(0). The slope of the tangent line at 0 is -3 because f '(x) =-3. </p>
<p>For part B:
You need to use the second derivative to find the points of inflection. The second derivative at 0 = 0, which tells you that it is not a point of inflection. If it were a max it would f " (x) would be positive. If it were a min f " (x) would be negative.</p>
<p>Another way you could do this is find critical values from your equation in part A. The only critical value is at x= 2/3, so f (0) is not a point of inflection. </p>
<p>I'll do the second half in one sec...</p>
<p>thanks a ton, u get the second half...?</p>
<p>For part C:
Can you solve g'(0) to find the slope at 0, and then add the intercept. If this is correct than the equation of the tangent line at 0 is y = 8x + 4.</p>
<p>For part D:
Just use the product rule: f '(x) g(x) + g '(x) f(x)= d/dx f(x)g(x)
So: g " (x) = e^(-2x) d/dx(3f '(x)+2f "(x)) + -2e^(-2x) (3 f(x) +2 f '(x))
Simplify that and I think it should equal the answer they gave.</p>
<p>For the last part of D:
No, it doesn't have a maximum at 0. To prove this you could find the critical values which will tell you where the maximum and minimum are.
g '(x) = 8x +4
0 = 8x +4
-4 = 8x
x = -1/2
So, x = -1/2 is a critical value.
Next, you need to put -1/2 into g"(x) to test if it is a maximum, minimum, or neither. I don't have a calculator, but when I did I ended up with e^0 (-42/2 + 6/2 + 0) = 1 (-36/2) = -18.
Thus, g (x) has a maximum at x= -1/2</p>
<p>wait for part C</p>
<p>g'(0) = e ^ 0 (3(2) + 2 (-3)
comes out to 0...that doesnt make sense; where did your calculation of 8x come from</p>
<p>so should it be Y = 4 ?</p>
<p>You're right...I think it should be y=4. I'm not sure where I got the y=4 from. I was doing some other homework at the time and got a little distracted. Sorry about that...It should be close to the same procedure.</p>