Question

question 8
this is what we know so far:
function: ( f(x)=sqrt3{x^{2}}-6 x+5 )
domain: ( (-infty, infty) )
intercepts: ( (0, sqrt3{5}) ) and ( (1,0) ) and ( (5,0) )
asymptotes: none
derivative: ( f^{prime}(x)=\frac{2 x-6}{3left(x^{2}-6 x+5
ight)^{\frac{2}{3}}} )
critical points: ( x=1, x=3 ), and ( x=5 )
decreasing on ( (-infty, 3) ). increasing on ( (3, infty) ).
local minimum: ( f(3)=-sqrt3{4} )
second derivative: ( f^{prime prime}(x)=-\frac{2 x^{2}-12 x+42}{9left(x^{2}-6 x+5
ight)^{\frac{5}{3}}} )
how many points are there where ( f^{prime prime}(x)=0 ) or ( f^{prime prime}(x) ) dne?
o 1
o 4
o none
o 2
o 3

Explanation:

Step1: Set the numerator of \(f''(x)\) equal to zero

We have \(f''(x)=-\frac{2x^{2}-12x + 42}{9(x^{2}-6x + 5)^{\frac{5}{3}}}\). Set the numerator \(2x^{2}-12x + 42=0\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 2\), \(b=-12\), \(c = 42\)), \(\Delta=b^{2}-4ac=(-12)^{2}-4\times2\times42=144 - 336=-192<0\). So, no real - valued solutions from the numerator.

Step2: Set the denominator of \(f''(x)\) equal to zero

Set \(x^{2}-6x + 5 = 0\). Factor the quadratic: \(x^{2}-6x + 5=(x - 1)(x - 5)=0\). Solving \((x - 1)(x - 5)=0\) gives \(x = 1\) and \(x = 5\). But the second derivative \(f''(x)\) is not defined at \(x = 1\) and \(x = 5\) (since the denominator is zero at these points).

Answer:

\(2\)