319HW10Solutions

# 319HW10Solutions - MAT 319/320 Solutions for HW10 Exercise...

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Unformatted text preview: MAT 319/320 Solutions for HW10 Exercise 1. Section 6.1, #1a. Proof. As usual we have to go back to the definition, so we write the quotient ( x + h ) 3 − x 3 h = x 3 + 3 x 2 h + 3 xh 2 + h 3 − x 3 h = 3 x 2 + h. (3 x + h 2 ) → 3 x 2 when h → square Exercise 2. Section 6.1, #4. Proof. The quotient f ( h ) − f (0) h = f ( h ) h is either equal to 0 or to h 2 h = h , thus in any case we have lessorequalslant vextendsingle vextendsingle vextendsingle f ( h ) − f (0) h vextendsingle vextendsingle vextendsingle lessorequalslant vextendsingle vextendsingle vextendsingle h vextendsingle vextendsingle vextendsingle → when h → , so by the squeeze theorem we deduce that f ′ (0) exists and is equal to zero. square Exercise 3. Section 6.1, #9. Proof. Assume that f is differentiable everywhere and that we have for any x ∈ R f (- x ) = f ( x ) . By taking the derivative of both sides we get (using the Chain rule for the left hand side): f ′ (- x ) . (- 1) = f ′ ( x ) thus f ′...
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319HW10Solutions - MAT 319/320 Solutions for HW10 Exercise...

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