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Maths-

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Question

The function $f open parentheses x close parentheses equals not stretchy integral from negative 1 to x of t open parentheses e to the power of t end exponent minus 1 close parentheses open parentheses t minus 1 close parentheses open parentheses t minus 2 close parentheses to the power of 3 end exponent open parentheses t minus 5 close parentheses to the power of 5 end exponent d t$ has local minimum at $x equals$

0
1
2
3

The correct answer is: 3

We have,
$f open parentheses x close parentheses equals not stretchy integral from negative 1 to x of t open parentheses e to the power of t end exponent minus 1 close parentheses open parentheses t minus 1 close parentheses open parentheses t minus 2 close parentheses to the power of 3 end exponent open parentheses t minus 3 close parentheses to the power of 5 end exponent d t$

$rightwards double arrow f to the power of ´ end exponent open parentheses x close parentheses equals x open parentheses e to the power of x end exponent minus 1 close parentheses open parentheses x minus 1 close parentheses open parentheses x minus 2 close parentheses to the power of 3 end exponent open parentheses x minus 3 close parentheses to the power of 5 end exponent$

Now, $f to the power of ´ end exponent open parentheses x close parentheses equals 0 rightwards double arrow x equals 0 comma blank 1 comma blank 2 comma blank 3$

Clearly, $f ´ left parenthesis x right parenthesis$ changes its sign from negative to positive in the neighbourhood of $x equals 3$

So, $x equals 3$ is a point of local minimum

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