<p>can somebody please help me with this problem?</p>
<p>A particle is moving along the x-axis with velocity v(t)= t sin t. Is the speed of the particle increasing or decreasing at t=2? explain fully.</p>
<p>thanks!!!</p>
<p>Ah **** I forgot this.</p>
<p>v’(2) = 0.77</p>
<p>Since it is positive, I am going to guess it is increasing.</p>
<p>i’m not trying to question you but are you pretty positive about your answer? (b/c i think i need to know it for tomorrow’s test! lol)</p>
<p>can you or somebody else please help me with 2 other questions?</p>
<p>1a. Let f be a twice-differentiable function such that f(2)=5 and f(5)=2. Let g be the function given by g(x)=f(f(x)). Explain why there must be a value c for 2<c<5 such that g’(c)=1</p>
<p>1b. Show that g’(2)=g’(5). Use this result to explain why there must be a value k for 2<k<5 such that g"(k)=0. (hint: use the Mean value theorem)</p>
<p>thanks so much! i appreciate your help! :)</p>
<p>1a) The Intermediate Value Theorem says so. That’s all there really is to it.
1b) Find g’ using chain rule. Prove g’2=g’5 by plugging in. Then use Rolle’s Theorem (the special case of Mean Value) to prove it. Think about a string connected by 2 tacks that are lined up horizontally. No matter how you twist and turn the string, there’s going to be at least 1 spot between the tacks where there is a min/max (horizontal tangent) so g’’ (in reality, g’) is 0.</p>
<p>1a. I’m sorry but i don’t get how the intermediate value theorm fits in with g(x) and f(f(x)).</p>
<p>1b. sorry to bother you once again but do you think you can solve the problem out? b/c i don’t really know where to go from your explanations…</p>
<p>but thanks for responding tho</p>
<p>please…can somebody help me with this once more?</p>
<p>1a. Let f be a twice-differentiable function such that f(2)=5 and f(5)=2. Let g be the function given by g(x)=f(f(x)). Explain why there must be a value c for 2<c<5 such that g’(c)=1</p>
<p>because g(5)=f(f(5))=5 and g(2)=f(f(2))=2,
the average rate of change ((f(b)-f(a))/(b-a)) is equal to:</p>
<p>(5-2)/(5-2)=1 and by the MVT (function is continuous and differentiable) there must be a value g’(c) equals to the Average rate of change, which is 1</p>
<p>1b. Show that g’(2)=g’(5). Use this result to explain why there must be a value k for 2<k<5 such that g"(k)=0. (hint: use the Mean value theorem)</p>
<p>since g’(x)=f’(f(x))<em>f’(x) then g’(2)=f’(f(2))</em>f’(2)=f’(5)<em>f’(2)
and g’(5)=f’(f(5))</em>f’(5)=f’(2)*f’(5)=g’(2)
since g’(5)=g’(2), by rolle’s theorem (if f(a)=f(b) then there must be a value f’(x)=0 between a and b equals 0) there must be a value g’'(k)=0 between 2 and 5</p>
<p>Columbia’s right.</p>
<p>I hope you were at least able to get the right answer in time for your test. It looks like you probably didn’t though.</p>