alleryFind the approximate value of $\sqrt{25.5}$ using differentials (given $\sqrt{25} = 5$):
A5.05
B5.5
C5.0498 (more precise)
D5.1
Answer & Solution
Correct answer: A. 5.05
Let $f(x) = \sqrt x$. $f'(x) = 1/(2\sqrt x)$.
$\Delta f \approx f'(25)\cdot 0.5 = (1/10)(0.5) = 0.05$.
**So $\sqrt{25.5} \approx 5 + 0.05 = 5.05$.**
Keep practising Unit 4-5: Applications of Derivatives & Curve Sketching
Browse all Unit 4-5: Applications of Derivatives & Curve Sketching questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
If the tangent to the curve $y = e^{2x}$ at point $(a, e^{2a})$ passes through origin, theIf $f(x) = x^x$ for $x > 0$, then $f$ has a minimum at $x = $:The function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on $[0, \infty)$ at:Find the **shortest distance** from origin to the curve $xy = 4$ (in the first quadrant):A rectangle is inscribed in a semicircle of radius $R$ with base on the diameter. The **maThe function $f(x) = x^3 - 3x + 1$ has critical points at:$f(x) = 2x^3 - 3x^2 - 12x + 5$. The **local maximum value** occurs at:If $f(x) = x^2 - 4x + 3$, the **interval** on which $f$ is decreasing is: