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Example: Evaluating the Function h(x)=2x+1h(x) = 2x + 1

To evaluate the linear function h(x)=2x+1h(x) = 2x + 1 for specific variables or algebraic expressions, substitute each expression for xx and simplify:

  • Evaluate h(k2)h(k^2): Replace xx with k2k^2 to get h(k2)=2(k2)+1h(k^2) = 2(k^2) + 1. Simplify to yield h(k2)=2k2+1h(k^2) = 2k^2 + 1.
  • Evaluate h(x+1)h(x+1): Replace xx with the binomial (x+1)(x+1) using parentheses to get h(x+1)=2(x+1)+1h(x+1) = 2(x+1) + 1. Distribute the 22 to obtain 2x+2+12x + 2 + 1. Simplify by combining the constant terms to find h(x+1)=2x+3h(x+1) = 2x + 3.
  • Evaluate h(x)+h(1)h(x) + h(1): This expression requires finding the numerical value of h(1)h(1) first and adding it to the algebraic expression for h(x)h(x). Find h(1)=2(1)+1=2+1=3h(1) = 2(1) + 1 = 2 + 1 = 3. Then, substitute both parts into the sum to get h(x)+h(1)=(2x+1)+(3)h(x) + h(1) = (2x + 1) + (3). Simplify by combining the constant terms to find h(x)+h(1)=2x+4h(x) + h(1) = 2x + 4.

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Updated 2026-05-06

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References

Tags

OpenStax

Intermediate Algebra @ OpenStax

Ch.3 Graphs and Functions - Intermediate Algebra @ OpenStax

Algebra

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