Assignment 1 for Math. 69.107ABCD,
Elementary Calculus I
Instructor: Dr. Angelo Mingarelli
Assignment 1
Math. 69.107A, Winter, 1997
Instructor: Prof. A. Mingarelli
Due: Jan. 21, 1997
NOTATION: Pi = 3.14159... is our usual Greek number
1. [10] Let f, g be differentiable functions with f '(0)=1, g(-1)=0 and g'(-1)=2.
Assume that the range of g is contained in the domain of f. What is the value
of the derivative of the composition fog at x=-1?
Hint: Use the Chain Rule...
Solution: 2 ... The Chain Rule says that (f o g)'(x) = f '(g(x)) g '(x) whenever f' exists at g(x) and g ' exists at x. In this case we are given that f '(g(-1)) g '(-1) = f '(0) g '(-1) = (1)(2) = 2.
2. [5] What is the derivative of the function f(x) = sqrt(x) sec(sqrt(x)) ?
Recall that "sqrt (x)" , means the square root of x.
Solution: (1/2) sec(sqrt(x)) tan(sqrt(x) + sec(sqrt(x))/(2 sqrt(x)). ... This is done using the Product Rule and the Chain Rule together.
3. [10] A curve is given by the equation x2 -2x +y2 = 9. Find the equation of the tangent line to the curve at the point (x,y) = (0,3).
Solution: y = (2/3)x + 3 ... OK, you need to use implicit differentiation, here. Recall that y is a function of x so we differentiate both sides of the equation: x2 -2x +y2 = 9. When we do this we get...2x + 2yy ' - 2 = 0, or, yy ' = 2 - x, or, solving for y ', we find y ' = (2 - x)/y. So, at x = 0, y '(0) = (2 - 0)/y(0) = 2/3... This is the slope of the tangent line at x = 0, right? Finally, we set y - yo = m (x - xo) with xo = 0 and yo = 3 and m = 2/3. This gives the required answer.
4. An excursion into "inverse functions".
A function f is defined by f (x) = (2x+3)/(2x-3) for x different from 3/2.
Another function F is defined by setting x = f (F(x)).
a) [5] What is the "explicit form" of the function F(x) ?
b) [5] What is the range of F(x) ? (i.e., the set of values F(x) as x varies over its domain...)
c) [5] Show that F'(x) = [ f '(F(x)) ]-1
Solution: a) F(x) = (3 + 2x)/(2x - 2) ...Why? We use the theory of inverse functions...i.e., we set x = f(F(x)) = (2F(x)+3)/(2F(x)-3) and now simply solve for the symbol "F(x)". We get, 2xF(x) - 3x = 2F(x) + 3 or factoring out F(x) and rearranging terms we get the form F(x) = (3 + 2x)/(2x - 2).
Solution: b) { x : x is real and x is not equal to 3/2} ... This is because Ran F(x) = Dom f(x) = { x : x is real and x is not equal to 3/2}, from the theory.
Solution: c) Use the Quotient Rule for the left-side, because you know F(x), from (a), above. Then find the right-side because you can find the quantities f '(F(x)) and its reciprocal explicitly. Finally, compare both sides...
Total: [40]