\(f(x) = (x^{2} + 3)^{2}\) is defines on the set of real numbers, R. Find the gradient of f(x) at x = \(\frac{1}{2}\).
Answer Details
To find the gradient of \(f(x) = (x^2+3)^2\) at \(x=\frac{1}{2}\), we need to differentiate f(x) with respect to x and then substitute x = \(\frac{1}{2}\). Using the chain rule, we have:
\begin{align*}
\frac{d}{dx}[(x^2+3)^2] &= 2(x^2+3) \cdot \frac{d}{dx}(x^2+3) \\
&= 2(x^2+3) \cdot 2x \\
&= 4x(x^2+3)
\end{align*}
So, the gradient of \(f(x)\) is \(4x(x^2+3)\). Substituting \(x=\frac{1}{2}\), we get:
\begin{align*}
\text{Gradient of } f(x) \text{ at } x = \frac{1}{2} &= 4\left(\frac{1}{2}\right)\left(\left(\frac{1}{2}\right)^2+3\right) \\
&= 4\left(\frac{1}{2}\right)\left(\frac{13}{4}\right) \\
&= \frac{13}{2} \\
&= 6.5
\end{align*}
Therefore, the gradient of \(f(x)\) at \(x=\frac{1}{2}\) is 6.5.