Question

the approximate value of $y = sqrt{4 + sin x}$ at $x = 0.12$, obtained from the tangent to the graph at $x = 0$ is
a
2
b
2.03
c
2.06
d
2.12
e
2.24

Explanation:

Step1: Find the derivative of the function

The function is \(y = \sqrt{4+\sin x}=(4 + \sin x)^{\frac{1}{2}}\).
Using the chain - rule \((u^n)^\prime=nu^{n - 1}\cdot u^\prime\), where \(u = 4+\sin x\) and \(n=\frac{1}{2}\).
\(y^\prime=\frac{1}{2}(4+\sin x)^{-\frac{1}{2}}\cdot\cos x=\frac{\cos x}{2\sqrt{4+\sin x}}\)

Step2: Evaluate the function and its derivative at \(x = 0\)

When \(x = 0\):

• \(y(0)=\sqrt{4+\sin0}=\sqrt{4}=2\)
• \(y^\prime(0)=\frac{\cos0}{2\sqrt{4+\sin0}}=\frac{1}{2\times2}=\frac{1}{4}\)

Step3: Use the linear approximation formula

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

Answer:

\(2.03\) (Option B)