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in Derivatives by (28.3k points)
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Let f(x) = (x – a)2 + (x – b)2 + (x – c)2. Then, f(x) has a minimum at x = 

A. \(\frac{a+b+c}{3}\)

B. 3\(\sqrt {abc}\)

C. \(\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)

D. none of these

1 Answer

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Best answer

Option : (A)

f(x) = (x - a)2 +(x - b)2 +(x - c)2 

f'(x) = 2[x – a + x – b + x - c] 

f'(x) = 2[3x - a - b - c] 

f’(x) = 0 

2[3x - a - b - c] = 0 

3x - a - b- c = 0 

3x = a + b + c

x = \(\frac{a+b+c}{3}\)

Hence by second derivative test f’’(x) > 0 so it’s a point of minimum. 

f” (x) = 2(3(1)) 

= 6 > 0

So, 

x = \(\frac{a+b+c}{3}\) point of minimum.

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