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Find the location of the local extrema of the following function. \( f(x)=3-x^{\frac{2}{9}} \)

Ask by Park Warren. in the United States
Feb 27,2025

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The function \( f(x) = 3 - x^{\frac{2}{9}} \) does not have any local extrema.

Solución

The function \( f(x) = 3 - x^{\frac{2}{9}} \) does not have any local extrema. To find the local extrema of a function, we need to find the critical points by taking the derivative of the function and setting it equal to zero. Let's find the derivative of the function \( f(x) = 3 - x^{\frac{2}{9}} \) with respect to \( x \) and then set it equal to zero to find the critical points. Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(f^{\prime}\left(x\right)=\frac{d}{dx}\left(3-x^{\frac{2}{9}}\right)\) - step1: Use differentiation rules: \(f^{\prime}\left(x\right)=\frac{d}{dx}\left(3\right)-\frac{d}{dx}\left(x^{\frac{2}{9}}\right)\) - step2: Find the derivative: \(f^{\prime}\left(x\right)=0-\frac{2}{9}x^{-\frac{7}{9}}\) - step3: Remove 0: \(f^{\prime}\left(x\right)=-\frac{2}{9}x^{-\frac{7}{9}}\) - step4: Simplify: \(f^{\prime}\left(x\right)=-\frac{2\sqrt[9]{x^{2}}}{9x}\) The derivative of the function \( f(x) = 3 - x^{\frac{2}{9}} \) with respect to \( x \) is \( f'(x) = -\frac{2\sqrt[9]{x^{2}}}{9x} \). To find the critical points, we set the derivative equal to zero and solve for \( x \): \[ -\frac{2\sqrt[9]{x^{2}}}{9x} = 0 \] Solving this equation will give us the critical points of the function. Let's solve it. Solve the equation by following steps: - step0: Solve for \(x\): \(-\frac{2\sqrt[9]{x^{2}}}{9x}=0\) - step1: Find the domain: \(-\frac{2\sqrt[9]{x^{2}}}{9x}=0,x\neq 0\) - step2: Rewrite the expression: \(\frac{-2\sqrt[9]{x^{2}}}{9x}=0\) - step3: Cross multiply: \(-2\sqrt[9]{x^{2}}=9x\times 0\) - step4: Simplify the equation: \(-2\sqrt[9]{x^{2}}=0\) - step5: Change the sign: \(2\sqrt[9]{x^{2}}=0\) - step6: Rewrite the expression: \(\sqrt[9]{x^{2}}=0\) - step7: Set the radicand equal to 0: \(x^{2}=0\) - step8: Set the base equal to 0: \(x=0\) - step9: Check if the solution is in the defined range: \(x=0,x\neq 0\) - step10: Find the intersection: \(x \in \varnothing \) The solution to the equation \( -\frac{2\sqrt[9]{x^{2}}}{9x} = 0 \) is \( x \in \varnothing \), which means there are no real solutions for \( x \). Therefore, the function \( f(x) = 3 - x^{\frac{2}{9}} \) does not have any local extrema.

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Bonus Knowledge

To find the local extrema of the function \( f(x) = 3 - x^{\frac{2}{9}} \), we first need to determine its critical points by taking the derivative and setting it to zero. The derivative is: \[ f'(x) = -\frac{2}{9} x^{-\frac{7}{9}}. \] Setting the derivative equal to zero gives us: \[ -\frac{2}{9} x^{-\frac{7}{9}} = 0. \] However, this expression can never be zero, which means there are no critical points from this equation. The derivative is defined for all \( x > 0 \) and is undefined for \( x \leq 0 \). Thus, let's analyze the behavior of the function as \( x \) approaches 0 and as \( x \) increases: - As \( x \to 0^+ \), \( f(x) \to 3 \). - As \( x \to \infty \), \( f(x) \to -\infty \). In fact, since the derivative \( f'(x) \) is negative for \( x > 0 \), the function is decreasing throughout its domain. The only potential local extremum occurs at \( x = 0 \), where the function achieves its maximum value of \( f(0) = 3 \). So, the local extremum is at \( x = 0 \) with a maximum value of 3. In summary, the local maximum occurs at \( (0, 3) \).

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