Section 4.2

Rolle’s theorem: Suppose 1.) f continuous on [a, b] 2.) f differentiable on (a, b) 3.) f(a) = f(b).

Then there exists c ∈ (a, b) such that f′(c) = 0 Mean Value Theorem: Suppose 1.) f continuous on [a, b] 2.) f differentiable on (a, b)

Then there exists c ∈ (a, b) Ruth that f′(c) = \frac{f(b)−f(a)}{b−a}

Proof: Let h(x) = f(x)− secant line.

h(x) = Applications of MVT (including Rolle’s) Ex 1: Show x^{5} + 3x^{3} + 2x − 4 = 0 has exactly one real root. Step 1: Show there exists a root (review IVT)

Step 2: Show there is at most one root.

Ex 2: Suppose f(1) = 2 and f′(x) ≤ 3. How large can f(4) be?

Ex 3: If f′(x) = 0 for all x ∈ (a, b), then f(x) = c for some constant c.

Ex 4: If f′(x) = g′(x) for all x ∈ (a, b), then f(x) = g(x) + c for some constant c.

Section 4.4: Indeterminate forms:

“∞ − ∞”  “∞ · 0”  “\frac{∞}{∞}”  “\frac{0}{0}”
“∞^{0}”  “0^{0}”  “1^{∞}”

L’Hospital’s Rule: Suppose f′ and g′ exist, g′(x) ≠ 0 near a except possibly at a where a ∈ R ∪ {+∞, −∞} and if either 1.) lim_{x→a}f(x) = lim_{x→a}g(x) = 0 or 2.) lim_{x→a}f(x) = ±∞ and lim_{x→a}g(x) = ±∞ and if lim_{x→a} \frac{f′(x)}{g′(x)} exists, then

lim_{x→a} \frac{f(x)}{g(x)} = lim_{x→a} \frac{f′(x)}{g′(x)}

Ex 1) lim_{x→+∞} \frac{−2x^{2}+3}{5x^{2}+4x} Old method: lim_{x→+∞} \frac{−2x^{2}+3}{5x^{2}+4x} = lim_{x→+∞} \frac{x^{2}(−2+ \frac{3}{x^{2}} )}{x^{2}(5+ \frac{4}{x})} = New method: lim_{x→+∞} \frac{−2x^{2}+3}{5x^{2}+4x}

lim_{x→1} \frac{1−x+ln(x)}{x^{3}−3x+2}