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Evaluate the function at each specified value of the independent variable and simplify. (If an answer is undefined, enter UNDEFINED.)

f(x) = { (x² - 16, 1-2x², x 0
f(x)=. { ( 1 -2x^2 , X> 0

f(-4)=

f(0) =

f(5) =



Evaluate the function at each specified value of the independent variable and simplify-example-1
User Joao Paulo
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1 Answer

4 votes

Answer:


\textsf{(a)}\quad f(-4)=\boxed{0}


\textsf{(b)}\quad f(0)=\boxed{-16}


\textsf{(c)}\quad f(5)=\boxed{-49}

Explanation:

A piecewise function gives multiple sub-functions for different parts of its domain.

Given piecewise function:


f(x)=\begin{cases}x^2-16, \;\;x\leq0\\1-2x^2,\;\;x > 0\end{cases}

Therefore, if x is less than or equal to zero, we should use f(x) = x² - 16.

Similarly, if x is greater than zero, we should use f(x) = 1 - 2x².


\hrulefill

Part (a)

To find f(-4), substitute x = -4 into the function.

As -4 ≤ 0, use the first part of the function, f(x) = x² - 16.


\begin{aligned}f(-4)&=(-4)^2-16\\&=16-16\\&=0\end{aligned}

Therefore:


\large\boxed{f(-4) = 0}


\hrulefill

Part (b)

To find f(0), substitute x = 0 into the function.

As 0 ≤ 0, use the first part of the function, f(x) = x² - 16.


\begin{aligned}f(0)&=(0)^2-16\\&=0-16\\&=-16\end{aligned}

Therefore:


\large\boxed{f(0) = -16}


\hrulefill

Part (c)

To find f(5), substitute x = 5 into the function.

As 5 > 0, use the second part of the function, f(x) = 1 - 2x².


\begin{aligned}f(5)&=1-2(5)^2\\&=1-2(25)\\&=1-50\\&=-49\end{aligned}

Therefore:


\large\boxed{f(5) = -49}

User JorgeM
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