the approximate value of $y = sqrt{4+sin x}$ at $x = 0.12$, obtained from the tangent to the graph at $x =…

Explanation:

Step1: Find the derivative of $y$

Let $y = \sqrt{4+\sin x}=(4 + \sin x)^{\frac{1}{2}}$. Using the chain - rule, if $u = 4+\sin x$, then $y = u^{\frac{1}{2}}$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$, and the derivative of $u$ with respect to $x$ is $\frac{du}{dx}=\cos x$. So, $\frac{dy}{dx}=\frac{\cos x}{2\sqrt{4+\sin x}}$.

Step2: Evaluate $y$ and $\frac{dy}{dx}$ at $x = 0$

When $x = 0$, $y(0)=\sqrt{4+\sin(0)}=\sqrt{4}=2$. When $x = 0$, $\frac{dy}{dx}\big|_{x = 0}=\frac{\cos(0)}{2\sqrt{4+\sin(0)}}=\frac{1}{2\times2}=\frac{1}{4}$.

Step3: Use the linear - approximation formula

The linear - approximation formula is $L(x)=y(a)+y^{\prime}(a)(x - a)$. Here, $a = 0$ and $x=0.12$. So, $L(0.12)=y(0)+y^{\prime}(0)(0.12 - 0)$. Substitute $y(0) = 2$ and $y^{\prime}(0)=\frac{1}{4}$ into the formula: $L(0.12)=2+\frac{1}{4}\times0.12$. $L(0.12)=2 + 0.03=2.03$.

Answer:

B. 2.03