Letting $f ( x ) = \sqrt { x }$ and $x _ { 0 } = 25 ,$ the approximation of $\sqrt { 25.1 }$ by differentials is

A)5.1
B)5.05
C)5.01
D)4.99
E)4.95

Question

Letting $f ( x ) = \sqrt { x }$ and $x _ { 0 } = 25 ,$ the approximation of $\sqrt { 25.1 }$ by differentials is

A)5.1
B)5.05
C)5.01
D)4.99
E)4.95

Quizplus · Accepted Answer

The differential approximation formula is $dy \approx f'(x_0) \cdot dx$, where $dx = x - x_0$. Given $f(x) = \sqrt{x}$, we find $f'(x) = \frac{1}{2\sqrt{x}}$. With $x_0 = 25$, $f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{10}$. For $x = 25.1$, $dx = 25.1 - 25 = 0.1$. Thus, $dy \approx \frac{1}{10} \cdot 0.1 = 0.01$. Since $f(x_0) = \sqrt{25} = 5$, the approximation is $5 + 0.01 = 5.01$.

Letting f(x)=xf ( x ) = \sqrt { x }f(x)=x​ And x0=25,x _ { 0 } = 25 ,x0​=25,

Letting $f ( x ) = \sqrt { x }$ And $x _ { 0 } = 25 ,$