For $f(x)=3x^4-8x^3+12x^2-48x+25$ on $[0,3]$, the critical point insid…

Question

For $f(x)=3x^4-8x^3+12x^2-48x+25$ on $[0,3]$, the critical point inside the interval is

Accepted Answer

B. $x=2$

1. $f'(x)=12x^3-24x^2+24x-48=12(x^3-2x^2+2x-4)$.
2. Factor: $x^3-2x^2+2x-4=x^2(x-2)+2(x-2)=(x-2)(x^2+2)$.
3. $f'(x)=0 \Rightarrow x=2$ (since $x^2+2>0$ has no real root).
4. So the only interior critical point is $x=2$.
_Source: syllabus Mathematics Ch 6 "Application of Derivatives", p.28_

Answer

$x=1$

Answer

$x=3$

Answer

$x=0$

For $f(x)=3x^4-8x^3+12x^2-48x+25$ on $[0,3]$, the critical point inside the interval is

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