Math 1251 Exam 2 Solutions

1. We find the slope of the tangent line by implicit diff.

7y^6*y'+e^x*y+e^x*y'=1/2, so putting x=2 y=1 gives

7y'+(e^2)+e^2*y'=1/2, or y'=(1/2-e^2)/(7+e^2)=m.

The line is y-2=m(x-1).

2.(a) y'=1/(x^2+arctan(4x))*(2x+1/(1+(4x)^2)*4)

(b) Use Lhospitals rule: lim_{x->0} cos(x*e^(-50))*e^(-50)/sec^2(2x)*2=


(c) Exponential growth is much faster than polynomial or logarithmic growth, so we cross of all but -e^x in the numerator, and -e^x in the denominator, which gives 1 as the limit.

3.f'(x) =x^3-2*x=x*(x^2-2), so the possible local max/mins are x=0,\sqrt(2), and x=-\sqrt(2). Since f''(x)=3x^2-2, f''(0)=-2<0, so 0 is a local max. f''(\sqrt(2))=4>0, so \sqrt(2) is a local min.




f(3)=11.25, so x=3 is the abs max, and x=\sqrt(2) is the abs min.

4. Take d/dt of xy=1 to get dx/dt*y+x*dy/dt=0. Then put x=2 y=1/2, dx/dt=3 to find that dy/dt=-3/4.

5. (a) True. If f(x) has a local maximum at c, then f'(c)=e^(-c^2)=0, but the exp function is never 0. Another way to see this is that f'(x)>0 so that f(x) is always increasing, and could never have a local max.

(b) False. The graph of f(x) could be increasing always, for example f(x)=x^3 has f'(0)=0, but every negative c has f(c)<0, and positive d has f(d)>0, so they would never be equal.