Math 1251 Exam 1 Solutions

1. The derivative dy/dx=2-1/x^2, evaluated at x=1, is 1. So the slope of the tangent line at (1,3) is m=1, and the tangent line is y-3=1(x-1).

2.(a) dy/dx=45*(3x-2)^44*3 by the chain rule

(b) f'(x)=cos(4-1/x^(1/3))*(-1)*(-1/3)*x^(-4/3).

(c) f(x)=x*(1-x^2)^(1/2), so f'(x)=(1-x^2)^(1/2)+x*(1/2)*(1-x^2)^(-1/2)*(-2x).

(d) If t gets large, |t+2| is about t and large, while sin(t^2) is between -1 and 1. So the numerator is about t. In the denominator, -4t beats out 5, so it is about -4t. The quotient then is -1/4, which is the limit.

(e) Factoring the numerator, (x^2-3x)/(x-3)=x so if x approaches 3, the answer is 3.

3.(f(x+h)-f(x))/h is the slope of the secant line connecting two points on the graph: (x,f(x)) and (x+h, f(x+h)). As h approaches 0, this secant line approaches the tangent line, so the quotient approaches the slope of the tangent line.

4.(a) False. Take f(x)=x. then d/dx(1/(f(x))=d/dx(1/x)=d/dx(x^(-1))= -x^(-2) which is not equal to 1/f'(x)=1/1=1. The correct answer should be -f'(x)/f(x)^2.

(b) False. the condition that |f(x)|<=1 means that the graph of y=f(x) lies between the two parallel lines y=1 and y=-1. It says nothing about the slope of the tangent line to the graph, which could be very large if f(x) is very wiggly. For example, f(x)=sin(2x) has f'(0)=2.