A real valued function f:[4, infty) rightarro | vedclass

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

Question

(B) Given $f(x) = (x^2+x+1)^{(x^2-3x-4)}$.Taking the natural logarithm on both sides, we get $\ln(f(x)) = (x^2-3x-4) \ln(x^2+x+1)$.Differentiating wit

Vedclass · Accepted Answer

monotonically increasing function

Vedclass · Answer

(B) Given $f(x) = (x^2+x+1)^{(x^2-3x-4)}$.Taking the natural logarithm on both sides, we get $\ln(f(x)) = (x^2-3x-4) \ln(x^2+x+1)$.Differentiating with respect to $x$:$\frac{f'(x)}{f(x)} = (2x-3) \ln(x^2+x+1) + (x^2-3x-4) \cdot \frac{2x+1}{x^2+x+1}$.For $x \in [4, \infty)$, $x^2+x+1 > 0$ and $x^2-3x-4 = (x-4)(x+1) \ge 0$.Since $x \ge 4$, $(2x-3) > 0$ and $\ln(x^2+x+1) > 0$, and $(x^2-3x-4) \ge 0$ and $(2x+1) > 0$.Thus, $f'(x) = f(x) [ (2x-3) \ln(x^2+x+1) + \frac{(x^2-3x-4)(2x+1)}{x^2+x+1} ] > 0$ for all $x > 4$.Therefore, $f(x)$ is a monotonically increasing function.

$A$ real valued function $f:[4, \infty) \rightarrow R$ is defined as $f(x)=(x^2+x+1)^{(x^2-3x-4)}$, then $f$ is

Similar Questions

If $\log (1+x)-\frac{2x}{2+x}$ is increasing,then

Let $f$ be any function continuous on $[a, b]$ and twice differentiable on $(a, b)$. If for all $x \in (a, b)$,$f^{\prime}(x) > 0$ and $f^{\prime \prime}(x) < 0$,then for any $c \in (a, b)$,$\frac{f(c)-f(a)}{f(b)-f(c)}$ is greater than

The function $\frac{a\sin x + b\cos x}{c\sin x + d\cos x}$ is decreasing,if

For what values of $a$ is the function $f(x) = x^{2} + ax + 1$ increasing on the interval $[1, 2]$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module