QUIZ #3: CALCULUS 1A (Stankova) Wednesday, February 11, 2004 Section 10:00–11:00 (Voight)

Problem 1.

(a) Find the slope of the tangent to the parabola y = 1 − 3x + x^{2} at the point where x = a.

(b) Find the slope of the tangent lines at the points whose x-coordinates are (i) 0 and (ii)1.

Solution. By definition, we have

m = f′(a) = lim_{x→a} \frac{f(x) − f(a)}{x − a} = lim_{x→a} \frac{(1 − 3x + x^{2}) − (1 − 3a + a^{2})}{x − a}

= lim_{x→a} \frac{x^{2} − a^{2} − 3(x − a)}{x − a} = lim_{x→a} \frac{(x − a)(x + a − 3)}{x − a}

= lim_{x→a}(x + a − 3) = 2a − 3.

[Using the definition

f′(a) = lim_{h→a} \frac{f(a + h) − f(a)}{h}

also is correct and will give the correct answer.]

For part (b), we substitute: f′(0) = 2(0)−3 = −3 and f′(1) = 2(1)−3 = −1.

QUIZ #3: CALCULUS 1A (Stankova) Wednesday, February 11, 2004 Section 11:00–12:00 (Voight)

Problem 1. Prove that the equation

x^{4} − 3x^{2} = 13

has at least one real root. Please verify in writing all hypotheses and conditions of any theorem you apply and explain your work.

Solution. We choose f(x) = x^{4} − 3x^{2} − 13, and seek a c such that f(c) = N = 0.

Cooking up a and b, we see that f(0) = −13 < 0, so we take a = 0, and f(3) = 3^{4} − 3^{3} − 13 = 41 > 0, so we take b = 3.

The function f is continuous on [0, 3] (actually, on (−∞, ∞)) because it is a polynomial. Therefore, by the Intermediate Value theorem, since

f(0) = −13 < 0 < 41 = f(3),

there exists a c in (0, 3) such that f(c) = 0, i.e. c^{4} − 3c^{2} = 13 as desired.