Math  /  Algebra

Questionh(x)={5x17, for x<31, for 3x<1x+4, for x1h(10)=h(2)=h(1)=h(4)=\begin{array}{l}h(x)=\left\{\begin{array}{ll}-5 x-17, & \text { for } x<-3 \\ 1, & \text { for }-3 \leq x<1 \\ x+4, & \text { for } x \geq 1\end{array}\right. \\ h(-10)=\square \\ h(-2)=\square \\ h(1)=\square \\ h(4)=\square\end{array}

Studdy Solution

STEP 1

1. The function h(x) h(x) is a piecewise function with different expressions for different intervals of x x .
2. We need to evaluate h(x) h(x) at specific points: x=10 x = -10 , x=2 x = -2 , x=1 x = 1 , and x=4 x = 4 .

STEP 2

1. Identify the correct piece of the piecewise function for each given x x .
2. Evaluate the function at x=10 x = -10 .
3. Evaluate the function at x=2 x = -2 .
4. Evaluate the function at x=1 x = 1 .
5. Evaluate the function at x=4 x = 4 .

STEP 3

Identify the correct piece of the piecewise function for each given x x :
- For x<3 x < -3 , use h(x)=5x17 h(x) = -5x - 17 . - For 3x<1 -3 \leq x < 1 , use h(x)=1 h(x) = 1 . - For x1 x \geq 1 , use h(x)=x+4 h(x) = x + 4 .

STEP 4

Evaluate the function at x=10 x = -10 :
Since 10<3 -10 < -3 , we use h(x)=5x17 h(x) = -5x - 17 .
h(10)=5(10)17 h(-10) = -5(-10) - 17 h(10)=5017 h(-10) = 50 - 17 h(10)=33 h(-10) = 33

STEP 5

Evaluate the function at x=2 x = -2 :
Since 32<1 -3 \leq -2 < 1 , we use h(x)=1 h(x) = 1 .
h(2)=1 h(-2) = 1

STEP 6

Evaluate the function at x=1 x = 1 :
Since x=1 x = 1 and x1 x \geq 1 , we use h(x)=x+4 h(x) = x + 4 .
h(1)=1+4 h(1) = 1 + 4 h(1)=5 h(1) = 5

STEP 7

Evaluate the function at x=4 x = 4 :
Since x=4 x = 4 and x1 x \geq 1 , we use h(x)=x+4 h(x) = x + 4 .
h(4)=4+4 h(4) = 4 + 4 h(4)=8 h(4) = 8
The evaluated values are: h(10)=33 h(-10) = 33 h(2)=1 h(-2) = 1 h(1)=5 h(1) = 5 h(4)=8 h(4) = 8

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