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Chapter 1: Problem 3

Evaluate (if possible) the function at the given value(s) of the independent variable. Simplify the results. \(g(x)=3-x^{2}\) (a) \(g(0)\) (b) \(g(\sqrt{3})\) (c) \(g(-2)\) (d) \(g(t-1)\)

Short Answer

Expert verified
The evaluated results of the function for the given values of x are: (a) g(0) = 3, (b) g(鈭�3) = 0, (c) g(-2) = -1, (d) g(t-1) = -t^2 + 2t + 2.

Step by step solution

01

(a)

We substitute \(x\) with 0 in the function \(g(x) = 3 - x^2\), hence \( g(0) = 3 - (0)^2 = 3 - 0 = 3\).
02

(b)

In the same way, when \(x = \sqrt{3}\), we substitute it into the function to get \(g(\sqrt{3}) = 3 - (\sqrt{3})^2 = 3 - 3 = 0\).
03

(c)

If \(x = -2\), through substitution, we get \(g(-2) = 3 - (-2)^2 = 3 - 4 = -1\).
04

(d)

Finally, if \(x\) is a variable \(t - 1\), we substitute it in the function and simplify. Thus \(g(t - 1) = 3 - (t - 1)^2 = 3 - (t^2 - 2t + 1) = -t^2 + 2t + 2\).

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Most popular questions from this chapter

Use the position function \(s(t)=-4.9 t^{2}+150\), which gives the height (in meters) of an object that has fallen from a height of 150 meters. The velocity at time \(t=a\) seconds is given by \(\lim _{t \rightarrow a} \frac{s(a)-s(t)}{a-t}\). Find the velocity of the object when \(t=3\).

Prove that if \(f\) has an inverse function, then \(\left(f^{-1}\right)^{-1}=f\).

$$ \lim _{x \rightarrow 2} f(x)=3, \text { where } f(x)=\left\\{\begin{array}{ll} 3, & x \leq 2 \\ 0, & x>2 \end{array}\right. $$

Prove that if \(f\) is continuous and has no zeros on \([a, b],\) then either \(f(x)>0\) for all \(x\) in \([a, b]\) or \(f(x)<0\) for all \(x\) in \([a, b]\)

Describe how the functions \(f(x)=3+\llbracket x \rrbracket\) and \(g(x)=3-\llbracket-x \rrbracket\) differ.
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