The interval in which the function f(x) = 2x^ | vedclass

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

Question

(D) Given function is $f(x) = 2x^2 - \log x$ for $x > 0$. To find the interval where the function decreases,we find the derivative $f'(x)$. $f'(x) = \

Vedclass · Accepted Answer

$\left(0, \frac{1}{2}\right)$

Vedclass · Answer

(D) Given function is $f(x) = 2x^2 - \log x$ for $x > 0$. To find the interval where the function decreases,we find the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(2x^2 - \log x) = 4x - \frac{1}{x} = \frac{4x^2 - 1}{x}$. For the function to be decreasing,we set $f'(x) $\frac{4x^2 - 1}{x} Since $x > 0$,the denominator is always positive. Thus,we must have $4x^2 - 1 $4x^2 Taking the square root,we get $|x| Given the condition $x > 0$,the interval becomes $0 Therefore,the function decreases in the interval $\left(0, \frac{1}{2}\right)$.

The interval in which the function $f(x) = 2x^2 - \log x$,for $x > 0$ decreases,is

Similar Questions

$A$ nucleotide is composed of which subunits?

The degree of ionization of $0.10 \ M$ lactic acid is $4.0 \ \%$. The value of $K_a$ is

Human urine is usually acidic because:

In the following common emitter circuit,if $\beta = 100$,$V_{CE} = 7\,V$,$V_{BE} = 0\,V$ (negligible),and $R_C = 2\,k\Omega$,then the value of $I_B$ is.......$mA$.

The inverse of the point $(1, 2)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module