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Question

Let $f open parentheses x close parentheses equals x to the power of 3 end exponent$ use Mean value theorem to write $fraction numerator f open parentheses x plus h close parentheses minus f left parenthesis x right parenthesis over denominator h end fraction equals f to the power of ´ end exponent left parenthesis x plus theta h right parenthesis$ with $0 less than theta less than 1$ . If $x not equal to 0$ , then $stack lim with h rightwards arrow 0 below invisible function application theta$ is equal to

-1
-0.5
0.5
1

The correct answer is: 0.5

Given, $f open parentheses x close parentheses equals x to the power of 3 end exponent$
$therefore blank f open parentheses x plus h close parentheses equals open parentheses x plus h close parentheses to the power of 3 end exponent$

Now, $f open parentheses x close parentheses equals 3 x to the power of 2 end exponent$

$therefore blank f open parentheses x plus theta h close parentheses equals 3 left parenthesis x plus theta h right parenthesis to the power of 2 end exponent$

Given, $fraction numerator f open parentheses x plus h close parentheses minus f left parenthesis x right parenthesis over denominator h end fraction equals f to the power of ´ end exponent left parenthesis x plus theta h right parenthesis$

$rightwards double arrow blank fraction numerator open parentheses x plus h close parentheses to the power of 3 end exponent minus x to the power of 3 end exponent over denominator h end fraction equals 3 open parentheses x plus theta h close parentheses to the power of 2 end exponent$

$rightwards double arrow blank fraction numerator x to the power of 3 end exponent plus h to the power of 3 end exponent plus 3 x h open parentheses x plus h close parentheses minus x to the power of 3 end exponent over denominator h end fraction$

$equals 3 left parenthesis x to the power of 2 end exponent plus theta to the power of 2 end exponent h to the power of 2 end exponent plus 2 x theta h right parenthesis$

$rightwards double arrow blank h to the power of 2 end exponent plus 3 x to the power of 2 end exponent plus 3 x h equals 3 x to the power of 2 end exponent plus 3 theta to the power of 2 end exponent h to the power of 2 end exponent plus 6 x theta h$

$rightwards double arrow blank h plus 3 x equals 3 theta to the power of 2 end exponent h plus 6 x theta$

Taking limit on both sides, we get

$stack lim with h rightwards arrow 0 below invisible function application open parentheses h plus 3 x close parentheses equals stack lim with h rightwards arrow 0 below invisible function application left parenthesis 3 theta to the power of 2 end exponent h plus 6 x theta right parenthesis$

$rightwards double arrow blank 3 x equals 6 x blank stack lim with h rightwards arrow 0 below invisible function application theta$

$rightwards double arrow blank stack lim with h rightwards arrow 0 below invisible function application theta equals fraction numerator 1 over denominator 2 end fraction equals 0.5$

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