According to it's second derivative, the function has points of inflection given by:
Answer B: 1 and 2.467.
What are the points of inflection of a function?
The points of inflection of a function are the values of x for which:
[tex]f^{\prime\prime}(x) = 0[/tex]
In this problem, the second derivative is:
[tex]f^{\prime\prime}(x) = x^2\cos{(\sqrt{x})} - 2x\cos{(\sqrt{x})} +\cos{(\sqrt{x})}[/tex]
Hence:
[tex]x^2\cos{(\sqrt{x})} - 2x\cos{(\sqrt{x})} +\cos{(\sqrt{x})} = 0[/tex]
[tex](x^2 - 2x + 1)\cos{(\sqrt{x})} = 0[/tex]
[tex]x^2 - 2x + 1 = 0 \rightarrow (x - 1)^2 = 0 \rightarrow x = 1[/tex]
[tex]\cos{(\sqrt{x})} = 0[/tex]
[tex]\sqrt{x} = \frac{\pi}{2}[/tex]
[tex](\sqrt{x})^2 = \left(\frac{\pi}{2}\right)^2[/tex]
[tex]x = 2.467[/tex]
Hence option B is correct.
You can learn more about points of inflection at https://brainly.com/question/10352137