If f(x) = frac{1}{4x^2 + 2x + 1},then its max | vedclass

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

Question

(A) To find the maximum value of $f(x) = \frac{1}{4x^2 + 2x + 1}$,we first find the minimum value of the denominator $g(x) = 4x^2 + 2x + 1$.Since $g(x

Vedclass · Accepted Answer

$4/3$

Vedclass · Answer

(A) To find the maximum value of $f(x) = \frac{1}{4x^2 + 2x + 1}$,we first find the minimum value of the denominator $g(x) = 4x^2 + 2x + 1$.Since $g(x)$ is a quadratic expression of the form $ax^2 + bx + c$ with $a = 4 > 0$,its minimum value occurs at $x = -\frac{b}{2a} = -\frac{2}{2(4)} = -\frac{1}{4}$.The minimum value of the denominator is $g(-\frac{1}{4}) = 4(-\frac{1}{4})^2 + 2(-\frac{1}{4}) + 1 = 4(\frac{1}{16}) - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1-2+4}{4} = \frac{3}{4}$.Since $f(x) = \frac{1}{g(x)}$,the maximum value of $f(x)$ is the reciprocal of the minimum value of $g(x)$.Therefore,the maximum value is $\frac{1}{3/4} = \frac{4}{3}$.

If $f(x) = \frac{1}{4x^2 + 2x + 1}$,then its maximum value is

Similar Questions

The quadratic expression $(2x+1)^{2} - px + q \neq 0$ for any real $x$, if

Number of integral values of $a$ for which both roots of the quadratic equation $x^2 - (2a + 3)x + a^2 + 3a = 0$ lie in the interval $(0, 4)$ is:

The set of values of $a$ for which the inequality $x^2 - (a + 2)x - (a + 3) < 0$ is satisfied by at least one positive real $x$ is:

For what possible values of $a$ does $6$ lie between the roots of the equation $x^2 + 2(a - 3)x + 9 = 0$?

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+2ax+(3a+10) = 0$ such that $\alpha < 1 < \beta$. Then the set of all possible values of $a$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module