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′) by choosing h(x) = g(x) and bk = βk ckπ ℓ This tells us that ckπ ℓ βk = 2 ℓ Rℓ 0 g(x)sin kπx ℓ dx So we have a solution u(x,t) = X∞ k=1 sin kπ ℓ x αk cos ckπ ℓ t) βk sin ℓ t (8) with αk = 2 ℓ Z ℓ 0 f(x)sin kπx ℓ dx βk = 2 ckπ Z ℓ 0 g(x)sin kπx ℓ dx While the sum (8) can be very complicated Example 1 Determine the linear approximation for f (x) = 3√x f ( x) = x 3 at x = 8 x = 8 Use the linear approximation to approximate the value of 3√805 805 3 and 3√25 25 3 Show Solution Since this is just the tangent line there really isn't a whole lot to finding the linear approximation f ′ ( x) = 1 3 x − 2 3 = 1 3 3 √ x 2One is to check the continuity of f (x) at x=3, and the other is to check whether f (x) is differentiable there First, check that at x=3, f (x) is continuous It's easy to see that the limit from the left and right sides are both equal to 9, and f (3) = 9 Next, consider differentiability at x=3 This means checking that the limit from the 4 X Mannol Hypoid 4l Gear Oil Sae