Math, asked by rowei0318, 6 months ago

Given h(x). Determine the value of -2h(-1)+h(3)-4h(2)?
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Answers

Answered by BrainlyIAS
14

Given :

\sf h(x) = \begin{cases} \sf{\sqrt{3 - x}} & \sf x \leq -1 \\ \sf x^{2} - 5x + 2 &  \sf -1 < x \leq 2 \\ \sf 7x - 9 & \sf x > 2 \end{cases}

To Find :

-2 h(-1) + h(3) - 4h(2)

Solution :

Let us find the values of h(x) ,

★══════════════════════★

  • h(-1)

\to \sf \sqrt{3-(-1)}

\to \sqrt{4}

\to 2

  • h(3)

\to 7(3)-9

\to 21-9

\to 12

  • h(2)

\to  (2)^2-5(2)+2

\to 4-10+2

\to 6-10

\to -4

★══════════════════════★

:\implies \textsf{\textbf{-2h(-1)\ +\ h(3)\ -\ 4h(2)}}\ \; \pink{\bigstar}

:\implies \sf -2(2)+(12)-4(-4)

:\implies \sf -4+12+16

:\implies \sf -4+28

:\implies \textsf{\textbf{24}}\ \; \green{\bigstar}

Answered by mathdude500
2

\large\underline\blue{\bold{Given \:  Question :-  }}

\begin{gathered}\bf h(x) = \begin{cases} \bf{\sqrt{3 - x}} & \sf x \leq -1 \\ \sf x^{2} - 5x + 2 & \bf -1 < x \leq 2 \\ \bf 7x - 9 & \sf x > 2 \end{cases}\end{gathered}

\sf \:Determine \:  the \:  value  \: of -2h(-1)+h(3)-4h(2)?

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\huge{AηsωeR} ✍

─━─━─━─━─━─━─━─━─━─━─━─━─

\large\underline\red{\bold{Given :-  }}

\begin{gathered}\bf h(x) = \begin{cases} \bf{\sqrt{3 - x}} & \sf x \leq -1 \\ \bf x^{2} - 5x + 2 & \sf -1 < x \leq 2 \\ \bf 7x - 9 & \sf x > 2 \end{cases}\end{gathered}

─━─━─━─━─━─━─━─━─━─━─━─━─

\large\underline\purple{\bold{To \:  Find :-  }}

\bf \: -2h(-1) \: + \: h(3) \: - \: 4h(2)

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\large\underline\purple{\bold{Solution :-  }}

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\bf \:Step :- 1.

\bf \:  ⟼ Calculation  \: of \:  h(- 1)

\sf \:  ⟼As x = - 1, h(x) =  \sqrt{3 - x}

\bf\implies \:h( - 1) =  \sqrt{3 - ( - 1)}  =  \sqrt{4}  = 2

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\bf \:Step :- 2.

\bf \:  ⟼Calculation  \: of \:  h(3)

\sf \:  ⟼As x = 3, h(x) = 7x - 9

\bf\implies \:h(3) = 7 \times 3 - 9 = 21 - 9 = 12

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\bf \:Step :- 3.

\bf \:  ⟼ Calculation  \: of  \: h(2)

\sf \:  ⟼As x = 2, h(x) =  {x}^{2}  - 5x + 2

\bf\implies \:h(2) =  {2}^{2}  - 5 \times 2 + 2  =  - 4

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\bf \:Step :- 4.

\bf \:Now,  \: consider :  -2h(-1)+h(3)-4h(2)

☆On substituting the values, we get

\bf \: =  - 2(2) + 12 - 4( - 4)

\bf \: =   - 4 + 12 + 16

\bf \: = 24

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\large{\boxed{\boxed{\bf{Hence, -2h(-1)+h(3)-4h(2) = 24}}}}

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