Question

Find c so thatf'(c)=f( 6)- f(4)/6-4 where f(x)=√ x+2 and c€(4,6)​

Answer 1

SOLUTION

TO DETERMINE

The value of c so that

$\displaystyle \sf{f'(c) = \frac{f(6) - f(4)}{6 - 4} }$

Where

$\displaystyle \sf{f(x) = \sqrt{x + 2} \: \: \: and \: \: c \in \: (4, 6) }$

EVALUATION

Here the given function is

$\displaystyle \sf{f(x) = \sqrt{x + 2} }$

Now

$\displaystyle \sf{f'(x) = \frac{1}{2 \sqrt{x + 2} } }$

Thus we get

$\displaystyle \sf{f(6) = \sqrt{6 + 2} = \sqrt{8} }$

$\displaystyle \sf{f(4) = \sqrt{4 + 2} = \sqrt{6} }$

$\displaystyle \sf{f'(c) = \frac{1}{2 \sqrt{c + 2} } }$

Now it is given that

$\displaystyle \sf{f'(c) = \frac{f(6) - f(4)}{6 - 4} }$

$\displaystyle \sf{ \implies \: \frac{1}{2 \sqrt{c + 2} } = \frac{ \sqrt{8} - \sqrt{6} }{2} }$

$\displaystyle \sf{ \implies \: \frac{1}{ \sqrt{c + 2} } = \frac{ \sqrt{8} - \sqrt{6} }{1} }$

$\displaystyle \sf{ \implies \: \sqrt{c + 2} = \frac{ 1 }{\sqrt{8} - \sqrt{6}} }$

$\displaystyle \sf{ \implies \: \sqrt{c + 2} = \frac{ \sqrt{8} + \sqrt{6} }{8 - 6} }$

$\displaystyle \sf{ \implies \: \sqrt{c + 2} = \frac{ \sqrt{8} + \sqrt{6} }{2} }$

$\displaystyle \sf{ \implies \: c + 2 = \frac{ 8 + 6 + 2 \sqrt{48} }{4} }$

$\displaystyle \sf{ \implies \: c + 2 = \frac{ 14 + 8 \sqrt{3} }{4} }$

$\displaystyle \sf{ \implies \: c + 2 = \frac{ 7 + 4 \sqrt{3} }{2} }$

$\displaystyle \sf{ \implies \: c = \frac{ 7 + 4 \sqrt{3} }{2} - 2}$

$\displaystyle \sf{ \implies \: c = \frac{ 3+ 4 \sqrt{3} }{2} }$

Clearly

$\displaystyle \sf{ 4 < \frac{ 3+ 4 \sqrt{3} }{2} < 6}$

$\displaystyle \sf{ \implies \: 4 < c < 6 }$

Hence the required value of c is given by

$\boxed{ \: \: \displaystyle \tt{ \: c = \frac{ 3+ 4 \sqrt{3} }{2} } \: \: \: }$

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