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QuestionGiven the piecewise function f(x)f(x), find: a. f(−2)f(-2), b. f(−25)f\left(-\frac{2}{5}\right), c. f(4)f(4), d. xx if f(x)=−2.5f(x)=-2.5.

Studdy Solution

STEP 1

Assumptions1. The function f(x)f(x) is defined as a piecewise function with three different expressions for different intervals of xx. . We are asked to find the value of the function at specific points and also to find the value of xx for a given function value.

STEP 2

To find f(−2)f(-2), we need to identify which interval −2-2 falls in. According to the definition of the function, −2-2 falls in the interval −5≤x≤−2-5 \leq x \leq -2. So, we use the corresponding expression 2x−2x - to find the value of the function.
f(−2)=2(−2)−f(-2) =2(-2) -

STEP 3

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

STEP 4

To find f(−2)f\left(-\frac{2}{}\right), we need to identify which interval −2-\frac{2}{} falls in. According to the definition of the function, −2-\frac{2}{} falls in the interval −2<x≤2-2 < x \leq2. So, we use the corresponding expression x2x^{2} to find the value of the function.
f(−2)=(−2)2f\left(-\frac{2}{}\right) = \left(-\frac{2}{}\right)^{2}

STEP 5

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 6

To find f(4)f(4), we need to identify which interval 44 falls in. According to the definition of the function, 44 falls in the interval 2<x≤52 < x \leq5. So, we use the corresponding expression −32x+2-\frac{3}{2}x + \frac{}{2} to find the value of the function.
f(4)=−32(4)+2f(4) = -\frac{3}{2}(4) + \frac{}{2}

STEP 7

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 = -2.5

STEP 8

To find xx for f(x)=−2.5f(x) = -2.5, we need to solve the equation f(x)=−2.5f(x) = -2.5 for each interval and check if the solution lies within the interval.
For the first interval −5≤x≤−2-5 \leq x \leq -2, we solve the equation 2x−3=−2.52x -3 = -2.5.
2x−3=−2.52x -3 = -2.5

STEP 9

olve for xx in the equation 2x−3=−2.52x -3 = -2.5.
2x=−2.5+3=.52x = -2.5 +3 =.5x=.52=.25x = \frac{.5}{2} =.25Since .25.25 does not lie in the interval −5≤x≤−2-5 \leq x \leq -2, there is no solution in this interval.

STEP 10

For the second interval −2<x≤2-2 < x \leq2, we solve the equation x2=−2.5x^{2} = -2.5.
x2=−2.5x^{2} = -2.5Since the square of a real number is always non-negative, there is no solution in this interval.

STEP 11

For the third interval <x≤5 < x \leq5, we solve the equation −3x+7=−.5-\frac{3}{}x + \frac{7}{} = -.5.
−3x+7=−.5-\frac{3}{}x + \frac{7}{} = -.5

STEP 12

olve for xx in the equation −2x+72=−2.5-\frac{}{2}x + \frac{7}{2} = -2.5.
−2x=−2.5−72=−2.5−.5=−6-\frac{}{2}x = -2.5 - \frac{7}{2} = -2.5 -.5 = -6x=−6−2=4x = \frac{-6}{-\frac{}{2}} =4Since 44 lies in the interval 2<x≤52 < x \leq5, the solution is x=4x =4.
So, the solutions area. f(−2)=−7f(-2) = -7 b. f(−25)=425f\left(-\frac{2}{5}\right) = \frac{4}{25} c. f(4)=−2.5f(4) = -2.5 d. x=4x =4 when f(x)=−2.5f(x) = -2.5

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