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Chapter 1: Problem 28

Evaluate (if possible) the function at each specified value of the independent variable and simplify. \(q(t)=\left(2 t^{2}+3\right) / t^{2}\) (a) \(q(2)\) (b) \(q(0)\) (c) \(q(-x)\)

Short Answer

Expert verified
The value of \(q(2)\) is 2.75, \(q(0)\) is undefined and \(q(-x)\) equals to the original function itself \(\frac{(2x^{2}+3)}{x^{2}}\).

Step by step solution

01

Understand the Function

The function provided is \(q(t)=\frac{(2t^{2}+3)}{t^{2}}\). It's a ratio of a polynomial to a quadratic term in 't'.
02

Evaluate \(q(2)\)

To find the value of \(q(2)\), substitute \(t=2\) into the function, which results in \(q(2)=\frac{(2*2^{2}+3)}{2^{2}}=\frac{(2*4+3)}{4}=\frac{11}{4} = 2.75\).
03

Evaluate \(q(0)\)

To find the value of \(q(0)\), substitute \(t=0\) into the function. However, because there's a \(t^{2}\) in the denominator, the function is undefined at \(t=0\), thus \(q(0)\) is undefined.
04

Evaluate \(q(-x)\)

To find the value of \(q(-x)\), substitute \(t=-x\) into the function, which results in \(q(-x)=\frac{(2*(-x)^{2}+3)}{(-x)^{2}}=\frac{(2x^{2}+3)}{x^{2}}\). Note that this is the same function.

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