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To determine the values of [tex]\( x \)[/tex] for which the second derivative [tex]\( f''(x) \)[/tex] of the function [tex]\( f(x) = \frac{5}{2} x^2 e^x \)[/tex] is zero, let us follow these steps:
1. Compute the first derivative [tex]\( f'(x) \)[/tex]:
- We need to apply the product rule for the differentiation of [tex]\( f(x) \)[/tex]. The product rule states that [tex]\( (uv)' = u'v + uv' \)[/tex], where [tex]\( u = \frac{5}{2} x^2 \)[/tex] and [tex]\( v = e^x \)[/tex].
- The derivative of [tex]\( u = \frac{5}{2} x^2 \)[/tex] is [tex]\( u' = 5x \)[/tex].
- The derivative of [tex]\( v = e^x \)[/tex] is [tex]\( v' = e^x \)[/tex].
- Using the product rule:
[tex]\[ f'(x) = \left( \frac{5}{2} x^2 \right)' e^x + \frac{5}{2} x^2 (e^x)' = 5x e^x + \frac{5}{2} x^2 e^x = e^x \left( 5x + \frac{5}{2} x^2 \right) \][/tex]
[tex]\[ f'(x) = \frac{5}{2} x e^x (2 + x) \][/tex]
2. Compute the second derivative [tex]\( f''(x) \)[/tex]:
- Again, apply the product rule to [tex]\( f'(x) = \frac{5}{2} e^x x (2 + x) \)[/tex]. We treat this as a product of two functions:
- Let [tex]\( u = \frac{5}{2} x (2 + x) \)[/tex] and [tex]\( v = e^x \)[/tex].
- The derivative of [tex]\( u \)[/tex] is:
[tex]\[ u' = \frac{5}{2} (2 + x + x) = \frac{5}{2} (2 + 2x) = \frac{5}{2} \cdot 2 (1 + x) = 5 (1 + x) \][/tex]
- The derivative of [tex]\( v = e^x \)[/tex] is [tex]\( v' = e^x \)[/tex].
- Applying the product rule again:
[tex]\[ f''(x) = \left( \frac{5}{2} x (2 + x) \right)' e^x + \frac{5}{2} x (2 + x) \left( e^x \right)' \][/tex]
[tex]\[ f''(x) = 5 (1 + x) e^x + \frac{5}{2} x (2 + x) e^x = e^x \left( 5 (1 + x) + \frac{5}{2} x (2 + x) \right) \][/tex]
[tex]\[ f''(x) = e^x \left( 5 + 5x + \frac{5}{2}(2x + x^2) \right) = e^x \left( 5 + 5x + 5x + \frac{5}{2} x^2 \right) \][/tex]
[tex]\[ f''(x) = e^x \left( 5 + 10x + \frac{5}{2} x^2 \right) = e^x \left( \frac{10 + 20x + 5x^2}{2} \right) = \frac{5}{2} e^x (2 + 4x + x^2) \][/tex]
3. Determine when [tex]\( f''(x) = 0 \)[/tex]:
- Since [tex]\( e^x \)[/tex] is never zero, we only need to solve:
[tex]\[ 2 + 4x + x^2 = 0 \][/tex]
- This is a quadratic equation. Solving it using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = 2 \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm \sqrt{8}}{2} = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2} \][/tex]
- Therefore, the solutions are:
[tex]\[ x = -2 + \sqrt{2} \quad \text{and} \quad x = -2 - \sqrt{2} \][/tex]
Therefore, the solutions can be approximated to:
[tex]\[ x \approx -3.414 \quad \text{and} \quad x \approx -0.586 \][/tex]
Given these calculations, the value of [tex]\( x \)[/tex] for which the second derivative [tex]\( f''(x) \)[/tex] equals zero is approximately [tex]\( x \approx -3.414 \)[/tex] and [tex]\( x \approx -0.586 \)[/tex], which do not match any of the given multiple-choice options exactly. This might indicate either the choices were not meant to match exactly numerical solutions or an approximation discrepancy. Hence, the choices 0, [tex]\(\ln 5\)[/tex], [tex]\(e^5\)[/tex], and [tex]\(5e\)[/tex] are not correct as none of them approximate the obtained solutions.
1. Compute the first derivative [tex]\( f'(x) \)[/tex]:
- We need to apply the product rule for the differentiation of [tex]\( f(x) \)[/tex]. The product rule states that [tex]\( (uv)' = u'v + uv' \)[/tex], where [tex]\( u = \frac{5}{2} x^2 \)[/tex] and [tex]\( v = e^x \)[/tex].
- The derivative of [tex]\( u = \frac{5}{2} x^2 \)[/tex] is [tex]\( u' = 5x \)[/tex].
- The derivative of [tex]\( v = e^x \)[/tex] is [tex]\( v' = e^x \)[/tex].
- Using the product rule:
[tex]\[ f'(x) = \left( \frac{5}{2} x^2 \right)' e^x + \frac{5}{2} x^2 (e^x)' = 5x e^x + \frac{5}{2} x^2 e^x = e^x \left( 5x + \frac{5}{2} x^2 \right) \][/tex]
[tex]\[ f'(x) = \frac{5}{2} x e^x (2 + x) \][/tex]
2. Compute the second derivative [tex]\( f''(x) \)[/tex]:
- Again, apply the product rule to [tex]\( f'(x) = \frac{5}{2} e^x x (2 + x) \)[/tex]. We treat this as a product of two functions:
- Let [tex]\( u = \frac{5}{2} x (2 + x) \)[/tex] and [tex]\( v = e^x \)[/tex].
- The derivative of [tex]\( u \)[/tex] is:
[tex]\[ u' = \frac{5}{2} (2 + x + x) = \frac{5}{2} (2 + 2x) = \frac{5}{2} \cdot 2 (1 + x) = 5 (1 + x) \][/tex]
- The derivative of [tex]\( v = e^x \)[/tex] is [tex]\( v' = e^x \)[/tex].
- Applying the product rule again:
[tex]\[ f''(x) = \left( \frac{5}{2} x (2 + x) \right)' e^x + \frac{5}{2} x (2 + x) \left( e^x \right)' \][/tex]
[tex]\[ f''(x) = 5 (1 + x) e^x + \frac{5}{2} x (2 + x) e^x = e^x \left( 5 (1 + x) + \frac{5}{2} x (2 + x) \right) \][/tex]
[tex]\[ f''(x) = e^x \left( 5 + 5x + \frac{5}{2}(2x + x^2) \right) = e^x \left( 5 + 5x + 5x + \frac{5}{2} x^2 \right) \][/tex]
[tex]\[ f''(x) = e^x \left( 5 + 10x + \frac{5}{2} x^2 \right) = e^x \left( \frac{10 + 20x + 5x^2}{2} \right) = \frac{5}{2} e^x (2 + 4x + x^2) \][/tex]
3. Determine when [tex]\( f''(x) = 0 \)[/tex]:
- Since [tex]\( e^x \)[/tex] is never zero, we only need to solve:
[tex]\[ 2 + 4x + x^2 = 0 \][/tex]
- This is a quadratic equation. Solving it using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = 2 \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm \sqrt{8}}{2} = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2} \][/tex]
- Therefore, the solutions are:
[tex]\[ x = -2 + \sqrt{2} \quad \text{and} \quad x = -2 - \sqrt{2} \][/tex]
Therefore, the solutions can be approximated to:
[tex]\[ x \approx -3.414 \quad \text{and} \quad x \approx -0.586 \][/tex]
Given these calculations, the value of [tex]\( x \)[/tex] for which the second derivative [tex]\( f''(x) \)[/tex] equals zero is approximately [tex]\( x \approx -3.414 \)[/tex] and [tex]\( x \approx -0.586 \)[/tex], which do not match any of the given multiple-choice options exactly. This might indicate either the choices were not meant to match exactly numerical solutions or an approximation discrepancy. Hence, the choices 0, [tex]\(\ln 5\)[/tex], [tex]\(e^5\)[/tex], and [tex]\(5e\)[/tex] are not correct as none of them approximate the obtained solutions.
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