Let g(x) = f(x) + f(1 - x) and f''(x)

Question

(D) Given $g(x) = f(x) + f(1 - x)$.Differentiating with respect to $x$, we get $g'(x) = f'(x) - f'(1 - x)$.Since $f''(x) < 0$, $f'(x)$ is a strictly d

Vedclass · Accepted Answer

$g(x)$ decreases on $[1/2, 1]$ and $g(x)$ increases on $[0, 1/2]$

Vedclass · Answer

(D) Given $g(x) = f(x) + f(1 - x)$.Differentiating with respect to $x$, we get $g'(x) = f'(x) - f'(1 - x)$.Since $f''(x) For $x \in [0, 1/2]$, we have $x \le 1 - x$, which implies $f'(x) \ge f'(1 - x)$ because $f'(x)$ is decreasing.Thus, $g'(x) = f'(x) - f'(1 - x) \ge 0$ for $x \in [0, 1/2]$, so $g(x)$ is increasing on $[0, 1/2]$.For $x \in [1/2, 1]$, we have $x \ge 1 - x$, which implies $f'(x) \le f'(1 - x)$ because $f'(x)$ is decreasing.Thus, $g'(x) = f'(x) - f'(1 - x) \le 0$ for $x \in [1/2, 1]$, so $g(x)$ is decreasing on $[1/2, 1]$.Therefore, $g(x)$ increases on $[0, 1/2]$ and decreases on $[1/2, 1]$.

Let $g(x) = f(x) + f(1 - x)$ and $f''(x) < 0$ for $0 \le x \le 1$. Which of the following is true regarding the monotonicity of $g(x)$?

Similar Questions

If $f(x) = \frac{x}{x^2+1}$ is an increasing function,then the value of $x$ lies in:

Let $f(x) = \sin x$ and $g(x) = x$.
Statement-$1$: For $x \in (0, \infty)$,$f(x) \leq g(x)$.
Statement-$2$: For $x \in (0, \infty)$,$f(x) \leq 1$ but as $x \rightarrow \infty$,$g(x) \rightarrow \infty$.

Which of the following is an increasing function?

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

Let $\lambda^{}$ be the largest value of $\lambda$ for which the function $f_{\lambda}(x) = 4\lambda x^{3} - 36\lambda x^{2} + 36x + 48$ is increasing for all $x \in \mathbb{R}$. Then $f_{\lambda^{}}(1) + f_{\lambda^{*}}(-1)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module