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

Ask by Hilton Warner. in the United States
Feb 27,2025

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Answer

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

Solution

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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Extra Insights

To find the local extrema of the function \( f(x) = 3 - x^{\frac{2}{9}} \), we first need to determine the derivative \( f'(x) \). The derivative is given by: \[ f'(x) = -\frac{2}{9}x^{-\frac{7}{9}}. \] Setting the derivative \( f'(x) = 0 \) gives no real solutions since the expression will never equal zero for real \( x \) (it can only be zero if multiplied by zero, which isn't the case here). However, we will analyze the behavior at the endpoints and where the derivative is undefined. The derivative is undefined at \( x = 0 \). Evaluating the function at 0, we find: \[ f(0) = 3 - 0^{\frac{2}{9}} = 3. \] To confirm that this is a local maximum, we can test the intervals around \( x = 0 \): - For \( x < 0 \), \( f'(x) \) is positive because we are squaring a negative number, thus \( f(x) \) is increasing. - For \( x > 0 \), \( f'(x) \) is negative, indicating \( f(x) \) is decreasing. Therefore, \( x = 0 \) is a local maximum, and the location of the local extrema is at \( x = 0 \) with \( f(0) = 3 \). In conclusion, the local extremum occurs at \( x = 0 \) and is a maximum point where \( f(0) = 3 \).

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