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Evaluate f(−5), f(0), and f(1) for the piecewise defined function.

f(x) = x² + 1 if x < 0
x if x ≥ 0

User Halfwarp
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1 Answer

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Final answer:

To evaluate the piecewise function, use the appropriate part of the definition based on whether the input is less than or greater than 0. For f(-5), use f(x) = x² + 1 to get 26, for f(0), use f(x) = x to get 0, and for f(1), also use f(x) = x to get 1.

Step-by-step explanation:

The question asks to evaluate f(-5), f(0), and f(1) for a piecewise defined function given by:

For x less than 0: f(x) = x² + 1

For x greater than or equal to 0: f(x) = x

Evaluating f(-5): Since -5 is less than 0, we use the first part of the function definition.

f(-5) = (-5)² + 1 = 25 + 1 = 26

Evaluating f(0): 0 is equal to 0, so we use the second part of the function definition.

f(0) = 0 = 0

Evaluating f(1): Since 1 is greater than 0, we again use the second part of the function definition.

f(1) = 1 = 1

�(−5)=26,�(0)=0,�(1)=1

f(−5)=26,f(0)=0,f(1)=1

So, �(−5)=26

f(−5)=26, �(0)=0

f(0)=0, and �(1)=1

f(1)=1 are the correct evaluations for the given piecewise defined function. The function switches behavior at

�=0x=0, providing different expressions for

�<0x<0 and �≥0x≥0.

User Beniaminus
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