For a function $f(x)$ to have an inverse on a domain $[a, \infty)$, it must be strictly monotonic (either strictly increasing or strictly decreasing) throughout that domain.
We are given the function $f(x) = 2x^3 - 3x^2 + 6$. To determine its monotonicity, we first find its derivative, $f'(x)$.
$ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 + 6) = 6x^2 - 6x $Now, we find the critical points by setting the derivative to zero:
$ 6x^2 - 6x = 0 $ $ 6x(x - 1) = 0 $This gives us critical points at $x = 0$ and $x = 1$. We analyze the sign of $f'(x)$ in the intervals defined by these points:
The function $f(x)$ must be strictly monotonic on the domain $[a, \infty)$ for it to have an inverse.
The smallest value of '$a$' from the given options that satisfies $a > 1$ is $a=2$.
If $a=2$, the domain is $[2, \infty)$. On this interval, $f'(x) > 0$, so the function is strictly increasing and has an inverse.
We need to find the range $B = [f(a), \infty)$.
Calculate $f(a)$ for $a=2$:
$ f(2) = 2(2)^3 - 3(2)^2 + 6 $ $ f(2) = 2(8) - 3(4) + 6 $ $ f(2) = 16 - 12 + 6 $ $ f(2) = 10 $Therefore, the range is $B = [10, \infty)$.
The smallest value of '$a$' among the options for which $f(x)$ is strictly monotonic on $[a, \infty)$ and thus has an inverse is $a=2$. The corresponding range is $B = [10, \infty)$.
Let the function $f (x) = \frac{x}{3} + \frac{3}{x} + 3$, $x\neq0$ be strictly increasing in $(-\infty, a_1)\cup(a_2,\infty)$ and strictly decreasing in $(a_3, \alpha_4)\cup(a_4,a_5)$. Then $\sum_{i=1}^5 a_i^2$ is equal to
Let $f(x) = \begin{cases} (x-1)^\frac{1}{2-x}, & x > 1, x \ne 2 \\ k, & x = 2 \end{cases}$ The value of k for which f is continuous at x = 2 is :