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Chapter 14: Q. 1 (page 889)

For the function $f (x) = {x^{2}, i f x \leq 0 x + 1, i f 0 < x < 2 5 - x, i f 2 \leq x \leq 5$ find $f (0)$ and $f (2)$

Short Answer

Expert verified

The value is $2, 3$ respectively

Step by step solution

01

Given information

Given function $f (x) = {x^{2}, x \leq 0 x + 1, i f 0 < x < 2 5 - x, i f 2 \leq x \leq 5$

02

Check the values satisfying the limits and calculate

For $x = 0$ ,

$f (x) = x^{2}$

$f (0) = 0^{2} f (0) = 0$

For $x = 2$

role="math" localid="1646982263287" $f (x) = 5 - x$

role="math" localid="1646982273636" $f (2) = 5 - 2 f (2) = 3$

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Most popular questions from this chapter

In Problems 13–32, use the accompanying graph of $y = f (x)$ .

Find the x-intercept(s), if any, of $f$ .

Use a graphing utility to find the indicated limit rounded to two decimal places.

$\lim_{x \to 2} \frac{x^{3} - 2 x^{2} + 4 x - 8}{x^{2} + x - 6}$

In Problems 7– 42, find each limit algebraically

$\lim_{x \to 1} (5 x^{4} - 3 x^{2} + 6 x - 9)$

In Problem, graph each function. Use the graph to find the indicated limit, if it exists.

$\underset{x \to 0}{l i m} f (x), f (x) = \{\begin{cases} e^{x} & i f x > 0 \\ 1 - x & i f x \leq 0 \end{cases}$

In Problems 7– 42, find each limit algebraically

$\lim_{x \to 1} (- 3)$

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