The denominator of a fraction is 16 more than | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the numerator be $x$. Then the denominator is $x^2 + 16$.The fraction is $f(x) = \frac{x}{x^2 + 16}$.To find the extrema,we find the derivativ

Vedclass · Accepted Answer

$ - 1/8$

Vedclass · Answer

(B) Let the numerator be $x$. Then the denominator is $x^2 + 16$.The fraction is $f(x) = \frac{x}{x^2 + 16}$.To find the extrema,we find the derivative $f'(x)$:$f'(x) = \frac{(x^2 + 16)(1) - x(2x)}{(x^2 + 16)^2} = \frac{16 - x^2}{(x^2 + 16)^2}$.Setting $f'(x) = 0$ gives $16 - x^2 = 0$,so $x = 4$ or $x = -4$.We evaluate $f(x)$ at these critical points:For $x = 4$,$f(4) = \frac{4}{16 + 16} = \frac{4}{32} = \frac{1}{8}$.For $x = -4$,$f(-4) = \frac{-4}{16 + 16} = \frac{-4}{32} = -\frac{1}{8}$.Comparing the values,the least value is $-1/8$.

The denominator of a fraction is $16$ more than the square of the numerator. Find the least value of the fraction.

Similar Questions

If $S_1$ and $S_2$ are respectively the sets of local minimum and local maximum points of the function $f(x) = 9x^4 + 12x^3 - 36x^2 + 25, x \in R$,then

$A$ metal wire $108 \ m$ long is bent to form a rectangle. If the area of the rectangle is maximum,then its dimensions are

Find the local maximum and local minimum values of the function given by $f(x) = x^{3} - 6x^{2} + 9x + 15$.

The sum of the perimeter of a circle and a square is $k$,where $k$ is a constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle.

The minimum value of $\left( x^2 + \frac{250}{x} \right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module