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$s(t)= 3/(t+2)^2-6(t+2)+9$

For what value of t is the function s above undefined? I am unsure when a function in mathematics is undefined or even defined, The answer to this problem is 1 I am looking for a clear explanation simple to understand and in-depth about when functions are undefined, how to check and any methods to distinguish defined from undefined.

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    $\begingroup$ Welcome to Math Stack Exchange. Division by $0$ is undefined $\endgroup$ Apr 7, 2019 at 3:49
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    $\begingroup$ Did you mean $\dfrac3{(t+2)^2}-6(t+2)+9$ or $\dfrac3{(t+2)^2-6(t+2)+9}$? $\endgroup$ Apr 7, 2019 at 3:57
  • $\begingroup$ the numerator is 3 so the one on the right $\endgroup$ Apr 7, 2019 at 4:22

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The function $s(t)= 3/(t+2)^2-6(t+2)+9$ is undefined when $t=-2$, because division by $0$ is undefined. For another example, in the context of real numbers, a function involving a square root would be undefined when the argument of the square root is a negative number.

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  • $\begingroup$ Since the original post said $s(t)= 3/(t+2)^2-6(t+2)+9$ without parentheses around $3/(t+2)^2-6(t+2)+9,$ I took it to mean $s(t)= \dfrac 3{(t+2)^2}-6(t+2)+9$ $\endgroup$ Apr 7, 2019 at 4:13
  • $\begingroup$ $s(t)=\dfrac 3 {(t+2)^2-6(t+2)+9}$ would be undefined when $(t+2)^2-6(t+2)+9=0$, i.e., $(t+2-3)^2=0$, i.e., $(t-1)^2=0$, i.e., $t=1$ $\endgroup$ Apr 7, 2019 at 4:25
  • $\begingroup$ Could we still let $t=-2$ in the domain of the function? Or the domain must only includes the inputs for which the function is defined? $\endgroup$
    – user599310
    Aug 19, 2020 at 21:07
  • $\begingroup$ @user599310: A function should be defined on its entire domain $\endgroup$ Aug 19, 2020 at 21:11
  • $\begingroup$ So for the above function if someone writes the domain as a set that includes $x=2$ then is not function? $\endgroup$
    – user599310
    Aug 19, 2020 at 21:16
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In a function f(x), "x" is the domain. if there is a value of x where you can not work out f(x) it means that f(x) is undefined for that value of x. Let's analyze an example: f(x)=a/b This function is defined for every value of b (with b been a real number) different from zero, remember we can not divide by zero. In every function where the denominator is zero there is an undefinition. Another example: f(x)= square root of x, in this case the function is defined for zero and every positive value of "x", we can not work out a square root of negative numbers, at least no in domain of the real numbers. So, all you need to do is: Check the operations in the equation of the function: +, -, *, powers when the index is an integer; those always can be worked out. divisions: the values where the divisor is zero are undefinitions. even roots can not be worked out when the number inside the root is negative therefore, where the expression inside the root is smaller than zero we have undefinitions. Logarithm: in a logarithm based a of b where "a" is smaller than zero or "a" is equal to 1 or "b" is smaller than zero you can not work out the logarithm, therefore in those places we also have undefinitions.

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    $\begingroup$ Please provide additional details in your answer. As it's currently written, it's hard to understand your solution. $\endgroup$
    – Community Bot
    Sep 9, 2021 at 6:25

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