To find [tex]\( f(0) \)[/tex], we need to evaluate the given piecewise function at [tex]\( x = 0 \)[/tex].
The piecewise function is defined as:
[tex]\[ f(x) = \begin{cases}
5x - 3 & \text{if } -5 \leq x \leq 3 \\
x^3 - 3 & \text{if } 3 \leq x \leq 4
\end{cases} \][/tex]
Since [tex]\( 0 \)[/tex] falls within the interval [tex]\([-5, 3]\)[/tex], we use the first piece of the function, which is [tex]\( f(x) = 5x - 3 \)[/tex]:
[tex]\[ f(0) = 5(0) - 3 = 0 - 3 = -3 \][/tex]
Therefore, [tex]\( f(0) = -3 \)[/tex].
(b) [tex]\( f(1) = \)[/tex]
To find [tex]\( f(1) \)[/tex], we evaluate the function at [tex]\( x = 1 \)[/tex].
Since [tex]\( 1 \)[/tex] also falls within the interval [tex]\([-5, 3]\)[/tex], we use the first piece of the function [tex]\( f(x) = 5x - 3 \)[/tex]:
[tex]\[ f(1) = 5(1) - 3 = 5 - 3 = 2 \][/tex]
Therefore, [tex]\( f(1) = 2 \)[/tex].
(c) [tex]\( f(3) = \)[/tex]
To find [tex]\( f(3) \)[/tex], we evaluate the function at [tex]\( x = 3 \)[/tex].
Since [tex]\( 3 \)[/tex] is the boundary point between the two intervals but falls within both, it is evident that we use the definition for the interval [tex]\([-5, 3]\)[/tex], i.e., the first part [tex]\( f(x) = 5x - 3 \)[/tex]:
[tex]\[ f(3) = 5(3) - 3 = 15 - 3 = 12 \][/tex]
Therefore, [tex]\( f(3) = 12 \)[/tex].
(d) [tex]\( f(4) = \)[/tex]
To find [tex]\( f(4) \)[/tex], we evaluate the function at [tex]\( x = 4 \)[/tex].
Since [tex]\( 4 \)[/tex] falls within the interval [tex]\([3, 4]\)[/tex], we use the second piece of the function [tex]\( f(x) = x^3 - 3 \)[/tex]:
[tex]\[ f(4) = 4^3 - 3 = 64 - 3 = 61 \][/tex]
Therefore, [tex]\( f(4) = 61 \)[/tex].
In conclusion, the values are:
(a) [tex]\( f(0) = -3 \)[/tex]
(b) [tex]\( f(1) = 2 \)[/tex]
(c) [tex]\( f(3) = 12 \)[/tex]
(d) [tex]\( f(4) = 61 \)[/tex]
