Question

question 15 - 1 point use the first derivative test to find the location of all local extrema for the function given below. enter an exact answer. if there is more than one local maximum or local minimum, write each value of x separated by a comma. if a local maximum or local minimum does not occur on the function, enter ∅ in the appropriate box. $f(x)=\frac{9x}{6x^{2}+9}$ local maxima occur at x = . local minima occur at x = . provide your answer below:

Explanation:

Step1: Find the derivative of the function

We have \(f(x)=\frac{9x}{6x^{2}+9}\). Using the quotient - rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 9x\), \(u^\prime=9\), \(v = 6x^{2}+9\), \(v^\prime = 12x\). Then \(f^\prime(x)=\frac{9(6x^{2}+9)-9x(12x)}{(6x^{2}+9)^{2}}=\frac{54x^{2}+81 - 108x^{2}}{(6x^{2}+9)^{2}}=\frac{81 - 54x^{2}}{(6x^{2}+9)^{2}}\).

Step2: Set the derivative equal to zero

Set \(f^\prime(x)=0\), so \(\frac{81 - 54x^{2}}{(6x^{2}+9)^{2}}=0\). Since the denominator \((6x^{2}+9)^{2}>0\) for all real \(x\), we solve \(81 - 54x^{2}=0\). Rearranging gives \(54x^{2}=81\), then \(x^{2}=\frac{81}{54}=\frac{3}{2}\), and \(x=\pm\frac{\sqrt{6}}{2}\).

Step3: Use the First - Derivative Test

Choose test points in the intervals \((-\infty,-\frac{\sqrt{6}}{2})\), \((-\frac{\sqrt{6}}{2},\frac{\sqrt{6}}{2})\), and \((\frac{\sqrt{6}}{2},\infty)\).
For \(x = - 2\) (in \((-\infty,-\frac{\sqrt{6}}{2})\)), \(f^\prime(-2)=\frac{81-54\times4}{(6\times4 + 9)^{2}}=\frac{81 - 216}{(24 + 9)^{2}}<0\).
For \(x = 0\) (in \((-\frac{\sqrt{6}}{2},\frac{\sqrt{6}}{2})\)), \(f^\prime(0)=\frac{81}{9^{2}}>0\).
For \(x = 2\) (in \((\frac{\sqrt{6}}{2},\infty)\)), \(f^\prime(2)=\frac{81-54\times4}{(6\times4 + 9)^{2}}<0\).
Since \(f(x)\) changes from decreasing to increasing at \(x=-\frac{\sqrt{6}}{2}\), there is a local minimum at \(x =-\frac{\sqrt{6}}{2}\). Since \(f(x)\) changes from increasing to decreasing at \(x=\frac{\sqrt{6}}{2}\), there is a local maximum at \(x=\frac{\sqrt{6}}{2}\).

Answer:

Local maxima occur at \(x=\frac{\sqrt{6}}{2}\)
Local minima occur at \(x=-\frac{\sqrt{6}}{2}\)