Math

QuestionFind f(2)f(-2), f(25)f(-\frac{2}{5}), and f(4)f(4) for the piecewise function f(x)f(x) defined as: f(x)={2x35x2x22<x232x+722<x5f(x)=\begin{cases} 2x-3 & -5 \leq x \leq -2 \\ x^2 & -2 < x \leq 2 \\ -\frac{3}{2}x + \frac{7}{2} & 2 < x \leq 5 \end{cases}

Studdy Solution

STEP 1

Assumptions1. The function f(x)f(x) is a piecewise function defined as - f(x)=x3f(x) =x -3 for 5x-5 \leq x \leq - - f(x)=xf(x) = x^{} for <x- < x \leq - f(x)=3x+7f(x) = -\frac{3}{}x + \frac{7}{} for <x5 < x \leq5

STEP 2

To find f(2)f(-2), we need to determine which part of the piecewise function to use. Since 2-2 is in the interval 5x2-5 \leq x \leq -2, we use the function f(x)=2xf(x) =2x -.

STEP 3

Substitute 2-2 into the function f(x)=2x3f(x) =2x -3.
f(2)=2(2)3f(-2) =2(-2) -3

STEP 4

Calculate the value of f(2)f(-2).
f(2)=2(2)3=43=7f(-2) =2(-2) -3 = -4 -3 = -7

STEP 5

To find f(25)f\left(-\frac{2}{5}\right), we need to determine which part of the piecewise function to use. Since 25-\frac{2}{5} is in the interval 2<x2-2 < x \leq2, we use the function f(x)=x2f(x) = x^{2}.

STEP 6

Substitute 25-\frac{2}{5} into the function f(x)=x2f(x) = x^{2}.
f(25)=(25)2f\left(-\frac{2}{5}\right) = \left(-\frac{2}{5}\right)^{2}

STEP 7

Calculate the value of f(25)f\left(-\frac{2}{5}\right).
f(25)=(25)2=425f\left(-\frac{2}{5}\right) = \left(-\frac{2}{5}\right)^{2} = \frac{4}{25}

STEP 8

To find f(4)f(4), we need to determine which part of the piecewise function to use. Since 44 is in the interval 2<x52 < x \leq5, we use the function f(x)=32x+72f(x) = -\frac{3}{2}x + \frac{7}{2}.

STEP 9

Substitute 44 into the function f(x)=32x+72f(x) = -\frac{3}{2}x + \frac{7}{2}.
f(4)=32(4)+72f(4) = -\frac{3}{2}(4) + \frac{7}{2}

STEP 10

Calculate the value of f(4)f(4).
f(4)=32(4)+72=6+72=6+3.5=2.5f(4) = -\frac{3}{2}(4) + \frac{7}{2} = -6 + \frac{7}{2} = -6 +3.5 = -2.5So, f(2)=7f(-2) = -7, f(25)=425f\left(-\frac{2}{5}\right) = \frac{4}{25}, and f(4)=2.5f(4) = -2.5.

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