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# plz help

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Let

f(x) = -x + 3 if x <= 0

f(x) = 2x - 15 if x > 0

How many solutions does the equation f(f(x)) = 4 have?

Apr 16, 2022

#1
+1

We divide the problem into cases.

Case 1: f(x) > 0.

Then,

$$\begin{array}{rcl} f(f(x)) &=& 4\\ 2f(x) - 15 &=& 4\\ f(x) &=& \dfrac{19}2 \end{array}$$

Case 2: f(x) <= 0.

Then,

$$\begin{array}{rcl} f(f(x)) &=& 4\\ -f(x) + 3&=& 4\\ f(x) &=& -1 \end{array}$$

Then f(x) = 19/2 or f(x) = -1.

We split the problem into 2 cases again.

Case A: x > 0.

Then,

$$2x - 15 = \dfrac{19}2\text{ or }2x - 15 = -1\\ x = \dfrac{49}4\text{ or }x = 7$$

Case B: x <= 0.

Then,

$$-x+3= \dfrac{19}2\text{ or } -x+3 = -1\\ x = -\dfrac{13}2\text{ or }x = 4\text{ (rejected)}$$

Then, solution set = $$\left\{\dfrac{49}4, 7, -\dfrac{13}2\right\}$$. So there are a total of 3 solutions.

Apr 16, 2022

#1
+1

We divide the problem into cases.

Case 1: f(x) > 0.

Then,

$$\begin{array}{rcl} f(f(x)) &=& 4\\ 2f(x) - 15 &=& 4\\ f(x) &=& \dfrac{19}2 \end{array}$$

Case 2: f(x) <= 0.

Then,

$$\begin{array}{rcl} f(f(x)) &=& 4\\ -f(x) + 3&=& 4\\ f(x) &=& -1 \end{array}$$

Then f(x) = 19/2 or f(x) = -1.

We split the problem into 2 cases again.

Case A: x > 0.

Then,

$$2x - 15 = \dfrac{19}2\text{ or }2x - 15 = -1\\ x = \dfrac{49}4\text{ or }x = 7$$

Case B: x <= 0.

Then,

$$-x+3= \dfrac{19}2\text{ or } -x+3 = -1\\ x = -\dfrac{13}2\text{ or }x = 4\text{ (rejected)}$$

Then, solution set = $$\left\{\dfrac{49}4, 7, -\dfrac{13}2\right\}$$. So there are a total of 3 solutions.

MaxWong Apr 16, 2022