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16. Let \( h(x) \) be a function defined for all \( x \neq 0 \) such that \( h(4)=-3 \) and the derivative of \( h \) is given by \( h^{\prime}(x)=\frac{x^{2}-2}{x} \) for all \( x \neq 0 \). (a) Find equation of the tangent line to the graph \( y=h(x) \) at \( x=4 \). (2 points) (b) Find all values of \( x \) for which the graph of \( h \) has a horizontal tangent. (2 points) (c) Find \( h^{\prime \prime}(x) \). (2 points)

Ask by O'Quinn Norton. in the United States
Jan 23,2025

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(c) \( h''(x) = 1 + \frac{2}{x^2} \).

Solución

**Problem 16 (c): Find \( h''(x) \).** To find the second derivative \( h''(x) \), we start with the first derivative: \[ h'(x) = \frac{x^2 - 2}{x} = x - \frac{2}{x} \] Now, take the derivative of \( h'(x) \) with respect to \( x \): \[ h''(x) = \frac{d}{dx} \left( x \right) - \frac{d}{dx} \left( \frac{2}{x} \right) = 1 - \left( -\frac{2}{x^2} \right) = 1 + \frac{2}{x^2} \] **Answer:** (c) The second derivative is \( h''(x) = 1 + \dfrac{2}{\,x^{2}\,} \).

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To solve the problem, we’ll work through each part step-by-step. ### (a) Equation of the Tangent Line at \( x=4 \) We start by calculating the derivative \( h'(4) \): \[ h'(x) = \frac{x^{2}-2}{x} \] Substituting \( x=4 \): \[ h'(4) = \frac{4^2 - 2}{4} = \frac{16 - 2}{4} = \frac{14}{4} = \frac{7}{2} \] Now we have both the point where \( x=4 \) and the slope of the tangent: - Point: \( (4, h(4)) = (4, -3) \) - Slope: \( \frac{7}{2} \) Using the point-slope form of the line: \[ y - y_1 = m(x - x_1) \] Substituting in our values: \[ y - (-3) = \frac{7}{2}(x - 4) \] Simplifying this: \[ y + 3 = \frac{7}{2}x - 14 \] \[ y = \frac{7}{2}x - 17 \] So the equation of the tangent line is: \[ y = \frac{7}{2}x - 17 \] ### (b) Values of \( x \) for Horizontal Tangent A horizontal tangent occurs when the derivative \( h'(x) = 0 \): \[ h'(x) = \frac{x^{2} - 2}{x} = 0 \] This simplifies to: \[ x^{2} - 2 = 0 \] Solving for \( x \): \[ x^{2} = 2 \quad \Rightarrow \quad x = \pm\sqrt{2} \] Thus, the values of \( x \) for which the graph of \( h \) has a horizontal tangent are: \[ x = \sqrt{2} \quad \text{and} \quad x = -\sqrt{2} \] ### (c) Find \( h^{\prime \prime}(x) \) We start with the first derivative: \[ h'(x) = \frac{x^{2}-2}{x} \] To find \( h''(x) \), we will differentiate \( h'(x) \): Using the quotient rule \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = x^{2} - 2 \) and \( v = x \): 1. \( u' = 2x \) 2. \( v' = 1 \) Now applying the quotient rule: \[ h''(x) = \frac{(2x)(x) - (x^{2} - 2)(1)}{x^{2}} = \frac{2x^{2} - (x^{2} - 2)}{x^{2}} = \frac{2x^{2} - x^{2} + 2}{x^{2}} = \frac{x^{2} + 2}{x^{2}} \] Thus: \[ h''(x) = 1 + \frac{2}{x^{2}} \] This captures the second derivative of the function \( h(x) \).

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