MATHEMATICS 120 PROBLEM SET 10 - SOLUTIONS
1. If f (x) is a twice differentiable function such that f (3) = 1, f 0 (3) = 1.5, and −2x ≤
f 00 (x) ≤ −x for all x > 0, find the smallest interval you can that contains f (3.2).
We have f (3.2) = f (3) + f 0 (3) · 0.2 + E2(3.2) = 1 + 1.5 · 0.2 + E2(3.2) = 1.3 + E2 (3.2), where
E2 (3.2) = f 00 (c) · 0.22 /2 = 0.02f 00 (c) for some c between 3 and 3.2. For such c we have
f 00 (c) ≥ −2x ≥ −2 · 3.2 = −6.4 and f 00 (c) ≤ −x ≤ −3. Hence −0.128 ≤ E2 (3.2) ≤ −0.06
and
1.172 = 1.3 − 0.128 ≤ f (3.2) ≤ 1.24 = 1.3 − 0.06.
2. Prove that if f (x) is an even function such that f (n) (x) exists everywhere for all n, then
all its Taylor polynomials Pn (x) about x = 0 are also even.
Since f (x) is even, f (−x) = f (x) for all x. If we differentiate this equation k times,
we get that (−1)k f (k) (−x) = f (k) (x). In particular, if x = 0 and k is odd, we get that
−f (k) (0) = f (k) (0), hence f (k) (0) = 0 for all odd k. Therefore any Taylor polynomial
Pn (x) of f about x = 0 has the form
Pn (x) = f (0) +
f 00 (0) 2 f (4) (0)
x +
+ ...,
2!
4!
with only even powers of x present. But if k is even, then cxk is an even function, hence
Pn (x) is also an even function.
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