Questions: Find f'(x) and find the value(s) of x where f'(x)=0. f(x)=x/(x^2+225) f'(x)=1/(x^2+255)

Transcript text: Find $f^{\prime}(x)$ and find the value(s) of $x$ where $f^{\prime}(x)=0$. \[ \begin{array}{l} f(x)=\frac{x}{x^{2}+225} \\ f^{\prime}(x)=\frac{1}{x^{2}+255} \end{array} \]

Solution

Solution Steps

To find $ f^{\prime}(x) $ for the function $ f(x) = \frac{x}{x^2 + 225} $, we will use the quotient rule for differentiation. The quotient rule states that if you have a function $ f(x) = \frac{u(x)}{v(x)} $, then its derivative is given by $ f^{\prime}(x) = \frac{u^{\prime}(x)v(x) - u(x)v^{\prime}(x)}{(v(x))^2} $. Here, $ u(x) = x $ and $ v(x) = x^2 + 225 $. After finding $ f^{\prime}(x) $, we will solve for $ x $ where $ f^{\prime}(x) = 0 $.

Step 1: Find the Derivative

To find the derivative of the function $ f(x) = \frac{x}{x^2 + 225} $, we apply the quotient rule. The derivative is given by:

\[ f^{\prime}(x) = \frac{(1)(x^2 + 225) - (x)(2x)}{(x^2 + 225)^2} = \frac{x^2 + 225 - 2x^2}{(x^2 + 225)^2} = \frac{225 - x^2}{(x^2 + 225)^2} \]

Step 2: Set the Derivative to Zero

To find the critical points, we set the derivative equal to zero:

\[ f^{\prime}(x) = 0 \implies 225 - x^2 = 0 \]

Solving for $ x $, we get:

\[ x^2 = 225 \implies x = \pm 15 \]