Question
(a) Topic-6.14 Selecting Techniques for Antidifferentiation
(b) Topic-8.9 Volume with Disc Method: Revolving Around the \(x\)- or \(y\)-Axis
(c) Topic-8.9 Volume with Disc Method: Revolving Around the \(x\)- or \(y\)-Axis
(d) Topic-8.9 Volume with Disc Method: Revolving Around the \(x\)- or \(y\)-Axis
Functions f, g, and h are twice-differentiable functions with g(2) = h(2) = 4. The line \(y = 4+\frac{2}{3}(x-2)\) is tangent to both the graph of g at x = 2 and the graph of h at x = 2.
(a) Find h'(2).
(b) Let a be the function given by a(x) = 3x3h(x). Write an expression for a'(x). Find a'(2).
(c) The function h satisfies \(h(x)=\frac{x^{2}-4}{1-(f(x))^{3}}\) for x ≠ 2. It is known that \(\lim_{x\rightarrow 2}h(x)\) can be evaluated using L’Hospital’s Rule. Use \(\lim_{x\rightarrow 2}h(x)\) to find f(2) and f'(2 ). Show the work that leads to your answers.
(d) It is known that g(x) ≤ h(x) for 1 < x < 3. Let k be a function satisfying g(x) ≤ k(x) ≤ h(x) for 1 < x < 3. Is k continuous at x = 2 ? Justify your answer.
▶️Answer/Explanation
\(\textbf{6(a)} \quad h'(2) = \frac{2}{3}\)
\(\textbf{6(b)} \quad a'(x) = 9x^2h(x) + 3x^3h'(x)\)
\(a'(2) = 9 \cdot 2^2 h(2) + 3 \cdot 2^3 h'(2) = 36 \cdot 4 + 24 \cdot \frac{2}{3} = 160\)
\(\textbf{6(c)} \quad \text{Because } h \text{ is differentiable, } h \text{ is continuous, so } \lim_{x \to 2} h(x) = h(2) = 4.\)
\(\quad \text{Also, } \lim_{x \to 2} h(x) = \lim_{x \to 2} \frac{x^2 – 4}{2 – (f(x))^3}, \text{ so } \lim_{x \to 2} \frac{x^2 – 4}{2 – (f(x))^3} = 4.\)
\(\quad \text{Because } \lim_{x \to 2} (x^2 – 4) = 0, \text{ we must also have } \lim_{x \to 2} \big(1 – (f(x))^3\big) = 0.\)
\(\quad \text{Thus } \lim_{x \to 2} f(x) = 1.\)
\(\quad \text{Because } f \text{ is differentiable, } f \text{ is continuous, so } f(2) = \lim_{x \to 2} f(x) = 1.\)
\(\quad \text{Also, because } f \text{ is twice differentiable, } f’ \text{ is continuous, so } \lim_{x \to 2} f'(x) = f'(2) \text{ exists.}\)
\(\quad \text{Using L’Hospital’s Rule,}\)
\(\quad \lim_{x \to 2} \frac{x^2 – 4}{2 – (f(x))^3} = \lim_{x \to 2} \frac{2x}{-3(f(x))^2 f'(x)} = \frac{4}{-3(1)^2 \cdot f'(2)}.\)
\(\quad \text{Thus } f'(2) = -\frac{1}{3}.\)
\(\textbf{6(d)} \quad \text{Because } g \text{ and } h \text{ are differentiable, } g \text{ and } h \text{ are continuous, so }\)
\(\quad \lim_{x \to 2} g(x) = g(2) = 4 \text{ and } \lim_{x \to 2} h(x) = h(2) = 4.\)
\(\quad \text{Because } g(x) \leq k(x) \leq h(x) \text{ for } 1 < x < 3, \text{ it follows from the squeeze theorem that } \lim_{x \to 2} k(x) = 4.\)
\(\quad \text{Also, } g(2) = k(2) = h(2) = 4, \text{ so } k(2) = 4.\)
\(\quad \text{Thus } k \text{ is continuous at } x = 2.\)