Question

unit 1 lesson 3 mohanidbarc
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the functions are defined:
· (f(x)=-2x)
· (g(x)=2x + 5)
· (h(x)=x^{2}+3x - 10)
calculate the indicated function values:

1. (f(b^{2}))
2. (f(g(x)))
3. (g(b^{2}))
4. (h(f(x)))
5. (h(b^{2}))
6. (h(g(x)))

Explanation:

Step1: Find $f(b^{2})$

Substitute $x = b^{2}$ into $f(x)$. Since $f(x)=-2x$, then $f(b^{2})=-2b^{2}$.

Step2: Find $f(g(x))$

First, find $g(x) = 2x + 5$. Then substitute $g(x)$ into $f(x)$. So $f(g(x))=-2(2x + 5)=-4x-10$.

Step3: Find $g(b^{2})$

Substitute $x = b^{2}$ into $g(x)$. Since $g(x)=2x + 5$, then $g(b^{2})=2b^{2}+5$.

Step4: Find $h(f(x))$

First, find $f(x)=-2x$. Then substitute $f(x)$ into $h(x)$. So $h(f(x))=(-2x)^{2}+3(-2x)-10 = 4x^{2}-6x - 10$.

Step5: Find $h(b^{2})$

Substitute $x = b^{2}$ into $h(x)$. Since $h(x)=x^{2}+3x - 10$, then $h(b^{2})=(b^{2})^{2}+3b^{2}-10=b^{4}+3b^{2}-10$.

Step6: Find $h(g(x))$

First, find $g(x)=2x + 5$. Then substitute $g(x)$ into $h(x)$. So $h(g(x))=(2x + 5)^{2}+3(2x + 5)-10$. Expand $(2x + 5)^{2}=4x^{2}+20x + 25$, and $3(2x + 5)=6x+15$. Then $h(g(x))=4x^{2}+20x + 25+6x + 15-10=4x^{2}+26x + 30$.

Answer:

1. $-2b^{2}$
2. $-4x - 10$
3. $2b^{2}+5$
4. $4x^{2}-6x - 10$
5. $b^{4}+3b^{2}-10$
6. $4x^{2}+26x + 30$