Elementary Functions and Calculus I Math 131 (Sec 42), Autumn 2004 Practice Mid-term 2 Solutions

1. Recall the definitions of the following.

(a) That a function f : R → R has a limit l at a point x. Answer. We say lim_{y→x} f(y) = l if for every ε > 0 there exists a δ > 0 such that if |y − x| < δ then |f(y) − f(x)| < ε.

(b) That a function f : R → R be continuous at a point x. Answer. If lim_{y→x} f(y) exists and equals f(x) then f is continuous at x.

(c) That a function f : R → R be differentiable at a point x. Answer. The function f is differentiable at x if

f′(x) := lim_{y→x} \frac{f(y) − f(x)}{y − x}

exists and is not equal to ±∞.

2. Recall the following results (without proving them). State clearly any assumptions that are required for each result to hold.

(a) The intermediate value theorem.

Answer. If f : I → R, where I is an interval in R, is continuous, a, b ∈ I and M is between f(a) and f(b), then there exists a c such that f(c) = M.

(b) The product rule.

Answer. If f and g are functions, differentiable at c then fg is differentiable at c and

(fg)′(c) = f′(c)g(c) + f(c)g′(c).

(c) The quotient rule.

Answer. If f and g are functions differentiable at c and g(c) ≠ 0, then f/g is differentiable at c and

(\frac{f}{g} )′ (c) = \frac{g(c)f′(c) − f(c)g′(c)}{g^{2}(c)}.

(d) The chain rule.

Answer. If g is differentiable at c and f is differentiable at g(c) then f ∘g is differentiable at c and (f ∘ g)(c) = f′(g(c))g′(c)

(e) The power rule. Answer. Let r be a rational number and define h : R → R by h(x) = x^{r} for all x ∈ R. Then write r = p/q where p and q have no common factors. If q is odd then h is differentiable at every x ∈ R. If q is even then h is differentiable at every x > 0. Whenever h is differentiable h′(x) = rx^{r−1}.

3. Apply the intermediate value theorem to show that p(x) = x^{7} + 10x^{3} + 2 has at least one root.

Answer. Observe p(−2) = −128 − 80 + 2 = −206 and p(1) = 13. Also we know that polynomials are continuous. Therefore by the intermediate value theorem (applied with M = 0) there exists a c such that p(c) = 0.

4. Use any of the definitions or theorems you have quoted above to find the derivative f′ of f when defined by the following formulae. State clearly which results you are using, and prove any steps which are not direct applications of the theorems.

(a) f(x) = 10x^{25} + 50x^{2} − 4x + 20x^{3} Answer. By the power rule and linearity of the derivative,

f′(x) = 250x^{2}4 + 100x − 4 + 60x^{2}.

(b) f(x) = (x^{3} + 40x)^{308} Answer. Let g(x) = x^{3} + 40x and h(x) = x^{308}, then by the power rule and linearity g′(x) = 3x^{2} + 40 and h′(x) = 308x^{307}. Thus

f′(x) = (h ∘ g)′(x) =^{chain rule} h′(g(x))g′(x) = 308(x^{3} + 40x)^{307}(3x^{3} + 40).

(c) f(x) = \frac{x^{9}+50}{x^{2}+4} Answer. Let g(x) = x^{9} + 50 and h(x) = x^{2} + 4, then by the power rule g′(x) = 9x^{8} and h′(x) = 2x. Thus

f′(x) = (\frac{g}{h})′ (x) =^{quotient rule} \frac{h(x)g′(x) − h′(x)g(x)}{h^{2}(x)} = \frac{9x^{8}(x^{2} + 4) − 2x(x^{9} + 50)}{(x^{2} + 4)^{2}}

5. Given that y : R → R is differentiable and satisfies x^{2}y(x) = 1 + y^{2}(x)x find y′(c) in terms of c and y(c).

Answer. Differentiating both sides of the equality gives

2xy(x) + x^{2}y′(x) = 2y(x)y′(x)x + y^{2}(x),

and so y′(x) = (y^{2}(x) − 2xy(x))/(x^{2} − 2y(x)x) provided the denominator is non-zero.

6. Let g : R → R be defined by g(x) = x(x + 1) if x ≤ 0 and g(x) = ax + b if x > 0 for some given a, b ∈ R. Find g′(y) for any y ≠ 0. For what values of a and b is g continuous at zero? For what values of a and b is g differentiable at zero? What must g′(0) be if it exists? Explain your answer.

Answer. If x < 0 then g(x) = x(x + 1) so by the product rule g′(x) = (x + 1) + x = 2x + 1. If x > 0 then g(x) = ax+b and so g′(x) = a. Observe g(0) = 0, lim_{x↘0} h(x) = lim_{x↘0} ax+b = b and lim_{x↗0} h(x) = lim_{x↗0} x(x + 1) = 0. Thus lim_{x→0} h(x) = h(0) if and only if the left-hand and right-hand limit exist and equal the value h(0), that is, if and only if b = 0. That is f is continuous at 0 exactly when b = 0 and a can be any number.

Now for g to be differentiable at 0 the limit lim_{x→0}(g(x) − b)/(x − 0) must exist. That is, the left-hand and the right-hand limits must exist and be equal. We can compute the right-hand limit is a and the left-hand limit is

lim_{x↗0} \frac{x(x + 1) − b}{x} = lim_{x↗0} \frac{x^{2} + x − b}{x} = lim_{x↗0} (x + 1 − \frac{b}{x})

which exists if and only if b = 0 (why?). When b = 0 the left-hand limit equals 1. Thus for the derivative to exist we must have a = 1 and b = 0.