Day 16: Functions, slopes, and limits

(a) Row by row, squaring first:
x = -2: x^2 = 4, f(-2) = 5
x = -1: x^2 = 1, f(-1) = 2
x = 0: x^2 = 0, f(0) = 1
x = 1: x^2 = 1, f(1) = 2
x = 2: x^2 = 4, f(2) = 5
Domain: every real x, because squaring and adding 1 are defined for all reals. Range: y ≥ 1. Reason: x^2 ≥ 0 with equality only at x = 0, so f(x) ≥ 1; and every y ≥ 1 is reached, since x = sqrt(y - 1) gives f(x) = y. The table shows only five outputs; the range claim needs this algebra.
One output per input: a function is a deterministic rule, so evaluating f(2) twice must give 5 both times. Two inputs may share an output (f(-2) = f(2) = 5); that is allowed. One input with two outputs is not.

(b) sqrt needs a non-negative argument, so x - 2 ≥ 0 and the domain is x ≥ 2. h(2) = sqrt(0) = 0, h(3) = sqrt(1) = 1, h(6) = sqrt(4) = 2. Range: y ≥ 0, because the principal square root is never negative and every y ≥ 0 is reached by x = y^2 + 2.

(c) {(1,2), (2,3), (1,4)} is not a function: input 1 is paired with both 2 and 4. x = y^2 does not define y as a function of x: x = 4 gives y = 2 and y = -2. Restricting to y ≥ 0 gives the function y = sqrt(x).

Common errors: writing (-2)^2 + 1 = -3; giving the range as all reals by copying the domain; calling f "not a function" because two inputs give 5.

(a) f(2) = 4 + 1 = 5 and f(5) = 25 + 1 = 26. Average rate of change = (26 - 5)/(5 - 2) = 21/3 = 7. The secant slope is the same number: the line through (2, 5) and (5, 26) rises 7 output units per input unit. Positive, as predicted.

(b) Point-slope form: y - 5 = 7(x - 2), so y = 7x - 9. Check: 7(5) - 9 = 26 = f(5), and 7(2) - 9 = 5 = f(2).

(c) f(2.5) = 7.25, slope (7.25 - 5)/0.5 = 4.5. f(2.1) = 5.41, slope 0.41/0.1 = 4.1. f(2.01) = 5.0401, slope 0.0401/0.01 = 4.01. The slopes approach 4.
Why: for this function the slope on [a, b] is ((b^2 + 1) - (a^2 + 1))/(b - a) = (b - a)(b + a)/(b - a) = a + b, valid because b != a. That explains every value: 2 + 5 = 7, 2 + 2.5 = 4.5, and a + b approaches 4 as b approaches 2. Day 17 calls that limit the derivative, f'(2) = 4.

(d) Units: metres per second. Over the 3 seconds from t = 2 to t = 5 the object covers 21 m, so its average speed is 7 m/s. This says nothing about the speed at any single instant.

A secant needs two distinct inputs: with b = a the quotient is 0/0 and undefined, which is why the instantaneous slope needs a limit.

(a) For x != 1, g(x) = (x - 1)(x + 1)/(x - 1) = x + 1. The cancellation is legal only because x - 1 != 0.
Table with the original formula: x = 0.9: (0.81 - 1)/(-0.1) = 1.9; x = 0.99: (0.9801 - 1)/(-0.01) = 1.99; x = 1.01: (1.0201 - 1)/0.01 = 2.01; x = 1.1: (1.21 - 1)/0.1 = 2.1. The simplified x + 1 gives the same four values.
From the left the outputs climb 1.9, 1.99 toward 2; from the right they fall 2.1, 2.01 toward 2. Both sides agree, so lim(x->1) g(x) = 2.
g(1) is not defined: the rule excludes x = 1. The graph is the line y = x + 1 with a hole at (1, 2). Writing g(1) = 2 would define a new function; it does not follow from the limit.

(b) k(x) = (x - 2)(x + 2)/(x - 2) = x + 2 for x != 2. k(1.9) = (3.61 - 4)/(-0.1) = 3.9 and k(2.1) = (4.41 - 4)/0.1 = 4.1, so lim(x->2) k(x) = 4, while k(2) is undefined.

(c) For positive x, |x|/x = x/x = 1; for negative x, |x|/x = -x/x = -1. So s(-0.1) = s(-0.01) = -1 and s(0.01) = s(0.1) = 1. The left limit is -1 and the right limit is 1. They differ, so lim(x->0) s(x) does not exist, and no factoring can repair it. Unlike (a), this is a jump, not a removable hole.

Common errors: substituting x = 1 into the original rule and writing 0/0 = 1; using values from one side only; claiming g(1) = 2.

(1) p(3) = 9 and p(3.1) = 3.1^2 = 9.61. Average slope = (9.61 - 9)/(3.1 - 3) = 0.61/0.1 = 6.1.

(2) With endpoint 3 + h: (p(3 + h) - p(3))/h = ((9 + 6h + h^2) - 9)/h = (6h + h^2)/h = 6 + h, valid for h != 0.
h = 0.1 gives 6.1 (matching (1)), h = 0.01 gives 6.01, h = 0.001 gives 6.001; from the left, h = -0.1 gives 5.9 and h = -0.01 gives 5.99. Both sides approach 6, so the secant slopes tend to 6, the instantaneous slope at 3.

(3) Assumptions: p is defined on an open interval around 3 (here, all reals); the quotient is formed only for h != 0; and the two-sided limit of the quotient exists, meaning the left and right values approach the same number. If the one-sided limits differed, as for |x| at 0, there would be no derivative.

(4) p(-1 + h) = 1 - 2h + h^2, so (p(-1 + h) - p(-1))/h = (-2h + h^2)/h = -2 + h for h != 0, which tends to -2. The slope at -1 is negative because the parabola is falling there. In general the limit is 2a, a pattern Day 17 proves.

Marking guide: dropping the h^2 term, cancelling h without stating h != 0, and checking only positive h are the three misses to record. Each points back to Activity 2 (secant quotient) or Activity 3 (limits).