MATH 180 FALL 1995

ASSIGNMENT #28

4.9, #1. f(x) = 1/x, f′(x) = −1/x^{2}. In particular, f(1) = 1, f′(1) = −1, so the local linearization at x = 1 is y = 1 − (x − 1) = 2 − x.

4.9, #6(a). f(x) = (1 + x)^{k}, f′(x) = k(1 + x)^{k−1}. In particular, f(0) = 1, f′(0) = k, so the local linearization at x = 0 is y = 1 + k(x − 0) = 1 + kx. 6(b). Part (a) for k = 1/2 gives the linearization y = 1 + .5x for \sqrt{1 + x}. Taking x = 0.1 then gives the estimate 1.05 for \sqrt{1.1} 6(c). The actual value is \sqrt{1.1} = 1.0488 up to four decimal places. The estimate in part (b) is larger than the actual value since f(x) is concave-down. Indeed, f′′(x) = −\frac{1}{4}(x + 1)^{−3/2} is always negative.

4.9, #9(a). 1/(1 + x) is the function of #6 with k = −1. So its linearization at x = 0 is 1 + kx = 1 − x. 9(b). When x is near 0, x^{2} is near 0 (in fact x^{2} is even nearer to 0). Substituting x^{2} for x in 1 − x ≈ 1/(1 + x) gives

1 − x^{2} ≈ \frac{1}{1 + x^{2}} .

9(c). The parabola 1 − x^{2} has derivative 0 at x = 0. Thus (1/(1 + x^{2})) has derivative 0 at x = 0.

4.Review, #28. The coefficient of h in the exponent of P = 30e^{−3.23×10^{−5}h} is negative. Thus the graph of P is one of exponential decay; the graph passes through (0, 30).

dP/dh|_{h=0} = 30(−3.23 × 10^{−5})e^{−3.23×10^{−5}h}|_{h=0} = −9.69 × 10^{−4}

So the tangent line at h = 0 is 30 − 9.69 × 10^{−4}h, the units being inches of mercury. The rule of thumb is the linear function P = 30 − 0.001h. This is esentially the equation of the tangent line since 9.69 × 10^{−4} = 0.000969. The approximation is less than the actual formula since P is concave-up.

4.Review, #32. f(x) = a^{x}, f′(x) = a^{x} ln a. The tangent line to a^{x} at x = 0 is 1 + (ln a)x since f(0) = 1 and f′(0) = ln a. The lines 1 + (ln a)x and x + 1 are the same only when a = e. If a = e, x + 1 meets the graph of a^{x} only at (0, 1) since a^{x} is concave-up. If a ≠ e, x + 1 meets the graph of a^{x} in two points. Indeed, if 1 < a < e, then x + 1 meets the graph in a second point to the right of (0, 1) since the slope of x + 1 is greater than the slope of 1 + (ln a)x. If e < a, then x + 1 meets the graph in a second point to the left of (0, 1) since x + 1 has smaller slope than the slope of 1 + (ln a)x.