2.5\nscore: 1/12 answered: 2/12\nquestion 3\nlet $f(x)=sqrt{4x^{2}+5x + 2}$\n$f(x)=\n$f(2)=\nquestion help…

Explanation:

Step1: Apply chain - rule

Let $u = 4x^{2}+5x + 2$, then $f(x)=\sqrt{u}=u^{\frac{1}{2}}$. The chain - rule states that $f^{\prime}(x)=\frac{df}{du}\cdot\frac{du}{dx}$. First, find $\frac{df}{du}$ and $\frac{du}{dx}$. $\frac{df}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=8x + 5$.

Step2: Substitute $u$ back

Substitute $u = 4x^{2}+5x + 2$ into $\frac{df}{du}$ and multiply by $\frac{du}{dx}$. So $f^{\prime}(x)=\frac{8x + 5}{2\sqrt{4x^{2}+5x + 2}}$.

Step3: Evaluate $f^{\prime}(2)$

Substitute $x = 2$ into $f^{\prime}(x)$. $f^{\prime}(2)=\frac{8\times2 + 5}{2\sqrt{4\times2^{2}+5\times2 + 2}}=\frac{16 + 5}{2\sqrt{16 + 10+2}}=\frac{21}{2\sqrt{28}}=\frac{21}{2\times2\sqrt{7}}=\frac{21}{4\sqrt{7}}=\frac{3\sqrt{7}}{4}$.

Answer:

$f^{\prime}(x)=\frac{8x + 5}{2\sqrt{4x^{2}+5x + 2}}$; $f^{\prime}(2)=\frac{3\sqrt{7}}{4}$