Recent content by teneleven

Homework Statement Find an equation of the tangent line to the curve at the given point. y = \frac{x - 1}{x - 2} , (3, 2)Homework Equations m = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h} y = mx + bThe Attempt at a Solution \lim_{h\to 0} \frac{\frac{(x + h) - 1}{(x + h) - 2} - \frac{x - 1}{x -...

Great, thanks for your help. I solved it using the method in my original post.Now I'm trying to solve it using the equation drpizza pointed out. f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}After plugging in y = -x^3 + 2x + 1 into the above equation I get... \lim_{h\to 0} \frac{-(h^3 + 3h^2x +...

Thanks for the replies. Factoring gets me... \lim_{x\rightarrow 1} -x^2 + 1 Which results in "0" Maybe I missed something. :/Using long division I get... m = \lim_{x\rightarrow 1} x^2 + 2x - 1 = 1 \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = 1 The point given is (1,2) so... y - 2 = 1(x - 1) y...

Homework Statement Find an equation of the tangent line to the curve at the given point. y = 1 + 2x - x^3, (1,2)Homework Equations y = mx + b m = \lim_{x\rightarrow a} \frac{f(x) - f(a)}{x - a}The Attempt at a Solution \lim_{x\rightarrow 1} \frac{(1 + 2x - x^3) - 2}{x - 1} =...

Why is ln wrong? If I have the following problem... \frac{1}{2}\int\frac{\sin\theta}{cos^2\theta}\ d\theta my first inclination is to solve it as follows: \frac{1}{2}\ln(cos^2\theta) + CFill me in.

What's wrong with having natural logs in the solution? How does one know to avoid ln when approaching this problem? I tried Dick's substitution of u = t^2 + 2 and didn't get very far. u = t^2 + 2 du = 2t\ dt \frac{1}{2}\int\frac{t^4}{\sqrt{t^2 + 2}}\ 2t Integration by parts...

Aye gents, I've been working on a problem that doesn't have a solution at the back of the book and would be grateful if you'd compare my answer with yours. Homework Statement \int\frac{t^5}{\sqrt{t^2 + 2}}\ dtHomework Equations t = \sqrt{2}\tan\theta dt = \sqrt{2}\sec^2\theta\ d\thetaThe...

Thanks for the prompt reply. I'm looking at it and I recall the property of logarithms, but I still don't understand how the "4" is reduced to \ln(4) and removed in the final answer. EDIT: So you're saying that D = C + \ln(4)?

I solved it using ChaoticLlama's method, but without integration by parts. I'm having trouble reducing it down to the answer in the book. I end up with a "4" in the denominator and haven't figured out how to get rid of it. \int\sec\theta\frac{(\sec\theta + \tan\theta)}{(\sec\theta +...

Homework Statement \int\frac{dx}{\sqrt{x^2 + 16}}Homework Equations x = 4\tan\theta dx = 4\sec^2\theta \ d\thetaThe Attempt at a Solution \int\frac{4\sec^2\theta}{\sqrt{16\tan^2\theta + 16}}\ d\theta \int\frac{4\sec^2\theta}{\sqrt{16(\tan^2\theta + 1)}}\ d\theta...

Thanks for the welcoming. Love the forums. I'm trying to simplify it but end up with something different. Where does the -18 come from? I've done the following: 9(\frac{\sqrt{x^2 + 9}}{3})^3 - \frac{27\sqrt{x^2 + 9}}{3} 9[\frac{(x^2 + 9)\sqrt{x^2 + 9}}{27}] - 9\sqrt{x^2 + 9} \frac{x^2 +...

Homework Statement \int\frac{x^3}{\sqrt{x^2 + 9}}Homework Equations x = 3\tan{\theta} dx=3\sec^2{\theta}The Attempt at a Solution 27\int\tan^3{\theta}\sec{\theta} 27\int\tan{\theta}(\sec^2{\theta} - 1)\sec{\theta} 27\int(sec^3{\theta} - \sec{\theta})\tan{\theta}...