question5 on Flickr - Photo Sharing!
The equation is at the link above.
For the 1st derivative I got f' = (49-x^2)^1/2 + (-x)(49-x^2)^-1/2
For the 2nd derivative I got f'' = -98x+x^3 / (49-x^2)^3/2
Are these right?
Thank you
question5 on Flickr - Photo Sharing!
The equation is at the link above.
For the 1st derivative I got f' = (49-x^2)^1/2 + (-x)(49-x^2)^-1/2
For the 2nd derivative I got f'' = -98x+x^3 / (49-x^2)^3/2
Are these right?
Thank you
Use product rule.. I think your answer has some errors:
$\displaystyle f(x) = x \times \sqrt{{49-x^{2}}}$
$\displaystyle f'(x) = [1 \sqrt{49-x^{2}}] + [x \times \frac{1}{2} \times {(49-x^{2})^\frac{-3}{2}} \times (-2x)]
= [\sqrt{49-x^{2}}] - [{x^2} \times {(49-x^{2})^\frac{-3}{2}}] $
$\displaystyle = \sqrt{49-x^{2}} - \frac{x^2}{\sqrt{49-x^{2}}} $
Now use the product rule to find the second derivative. I am a little too lazy topost the complete solution to find the second derivative. Try doing it yourself and post if you have problems. For your convenience, the second derivative looks like this:
$\displaystyle 2x \times [\frac{{-x}^2}{2(49-x^2)^\frac{3}{2}} -\frac{1}{2 \sqrt{49-x^{2}}}] - \frac{2x}{\sqrt{49-x^{2}}} $
Hello, and thank you for your responses!
I think I got the right f"
f"=2x^3-147x / (49-x^2)^3/2
I also think I got the inflection points x=0, x=-7, x=7
but now I am having a problem interpreting this into the right answer.
I know it can't be D. I'm pretty sure its not A or B because (0,7) is not on this function. So it is B???? Thank you!!!