Question

the second derivative of the function f is given by ( f(x) = x^2 cosleft( sqrt{x}
ight) - 2x cosleft( sqrt{x}
ight) + cosleft( sqrt{x}
ight) ). at what values of x in the interval (0, 3) does the graph of f have a point of inflection?
a 2.467 only
b 1 and 2.467
c 1.443 and 2.734
d 1 and 1.962

Explanation:

Step1: Recall Inflection Point Condition

A point of inflection occurs where the second derivative \( f''(x) \) changes sign (from positive to negative or vice - versa) and \( f''(x) = 0 \) (or is undefined, but here \( f''(x) \) is defined for \( x>0 \) in the interval \( (0,3) \)). So we first factor \( f''(x) \).

Given \( f''(x)=x^{2}\cos(\sqrt{x}) - 2x\cos(\sqrt{x})+\cos(\sqrt{x}) \), factor out \( \cos(\sqrt{x}) \):

\( f''(x)=\cos(\sqrt{x})(x^{2}-2x + 1) \)

Notice that \( x^{2}-2x + 1=(x - 1)^{2} \), so \( f''(x)=\cos(\sqrt{x})(x - 1)^{2} \)

Step2: Analyze the Factors

• The factor \( (x - 1)^{2}\geq0 \) for all real \( x \), and \( (x - 1)^{2}=0 \) when \( x = 1 \), and \( (x - 1)^{2}>0 \) for \( x

eq1 \) in the interval \( (0,3) \).

• The factor \( \cos(\sqrt{x}) \): we need to find when \( \cos(\sqrt{x}) = 0 \) and when its sign changes. The equation \( \cos(t)=0 \) when \( t=\frac{\pi}{2}+k\pi,k\in\mathbb{Z} \). Let \( t = \sqrt{x} \), so \( \sqrt{x}=\frac{\pi}{2}+k\pi \). For \( x\in(0,3) \), \( \sqrt{x}\in(0,\sqrt{3})\approx(0,1.732) \) when \( k = 0 \), \( t=\frac{\pi}{2}\approx1.571 \), \( x = (\frac{\pi}{2})^{2}\approx2.467 \); when \( k=- 1 \), \( t=-\frac{\pi}{2} \) (not in the domain of \( \sqrt{x} \) for \( x>0 \)).

Now, analyze the sign of \( f''(x) \):

• For \( x\in(0,1) \): \( (x - 1)^{2}>0 \), \( \cos(\sqrt{x})>0 \) (since \( \sqrt{x}\in(0,1)<\frac{\pi}{2} \)), so \( f''(x)>0 \).
• For \( x = 1 \): \( f''(1)=\cos(1)\times(1 - 1)^{2}=0 \), but \( (x - 1)^{2} \) does not change sign at \( x = 1 \) (it is a square), so the sign of \( f''(x) \) is determined by \( \cos(\sqrt{x}) \). Before \( x = 2.467 \), say \( x = 2 \), \( \sqrt{x}=\sqrt{2}\approx1.414<\frac{\pi}{2}\approx1.571 \), \( \cos(\sqrt{2})>0 \), so \( f''(2)>0 \). At \( x = 2.467 \), \( \cos(\sqrt{2.467})=\cos(\frac{\pi}{2}) = 0 \). After \( x = 2.467 \), say \( x = 3 \), \( \sqrt{3}\approx1.732>\frac{\pi}{2}\approx1.571 \), \( \cos(\sqrt{3})<0 \), and \( (x - 1)^{2}>0 \) for \( x = 3 \), so \( f''(3)<0 \). So \( f''(x) \) changes sign at \( x\approx2.467 \). At \( x = 1 \), since \( (x - 1)^{2} \) is non - negative and doesn't change sign, the sign of \( f''(x) \) is the same as the sign of \( \cos(\sqrt{x}) \) on both sides of \( x = 1 \) (because \( (x - 1)^{2} \) is positive on both sides of \( x = 1 \) in \( (0,3) \) except at \( x = 1 \) where it is zero). So \( x = 1 \) is not a point of inflection. We check the other options:
• Option A: \( x = 2.467 \) is a point of inflection as \( f''(x) \) changes sign there.
• Option B: \( x = 1 \) is not a point of inflection as explained.
• Option C: The values \( 1.443 \) and \( 2.734 \) do not satisfy the condition from the factored form of \( f''(x) \).
• Option D: \( x = 1 \) and \( x = 1.962 \): \( x = 1 \) is not a point of inflection, and \( 1.962 \) does not make \( f''(x) \) change sign.

Answer:

A. 2.467 only