f 2 (0) + (f ′ (0)) 2 = 85 f : R → [–2, 2]

This is true of every continuous function

f ′ =

|f ′ | =

–2 ≤ f(–4) ≤ 2

–2 ≤ f(0) ≤ 2

–4 ≤ f(-4) – f(0) ≤ 4

This |f ′ | ≤ 1

f(x) = 1

Note f(x) should have a bound ∞ which can be concluded by considering

f(x) = 2 sin

f ′ (x) = cos

f 2 (0) + (f ′ (0) 2 ) = 85

and does not exist

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Consider H(x) = f 2 (x) + (f ′ (x) 2

H(0) = 85

By choice there exists some x 0 such that (f ′ (x 0 )) 2 ≤ 1 for some x 0 in (-4, 0)

hence vr% H(x 0 ) = f 2 (x 0 ) + (f ′ (x 0 )) 2 ≤ 4 + 1

H(x 0 ) ≤ 5

Hence let p ∈ (-4, 0) for which H(p) = 5

(note that we have considered p as largest such negative number)

similarly let q be smallest positive number ∈ (0, 4) such that H(q) = 5

Hence By Rolle's theorem is (p, q)

H ′ = 0 for some c ∈ (–4, 4) and since H(x) is greater than 5 as we move from x = p

(f ′ (x)) 2 ≥ 1 in (p, q)

Thus vr% H ′ = 0 ⇒ f ′ f + f ′ f ′′ = 0

so y, f + f ′′ = 0 and f ′ ≠ 0