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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblema 4.9.9
Capitolo 4, Problema 4.9.9

If F(x) = x² - 3x + C and F (-1) = 4 , what is the value of C?

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Start with the given function F(x) = x² - 3x + C, where C is a constant to be determined.
Substitute x = -1 into the function F(x) since it is given that F(-1) = 4. This gives the equation: F(-1) = (-1)² - 3(-1) + C.
Simplify the terms in the equation: (-1)² = 1, -3(-1) = 3, so the equation becomes F(-1) = 1 + 3 + C.
Set F(-1) equal to 4 as given in the problem: 4 = 1 + 3 + C.
Solve for C by isolating it: Subtract 1 and 3 from both sides of the equation to find the value of C.

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Function Evaluation

Function evaluation involves substituting a specific input value into a function to determine its output. In this case, we need to evaluate the function F(x) at x = -1 to find the corresponding output, which is given as 4. This process is fundamental in calculus as it allows us to analyze the behavior of functions at specific points.
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Algebraic Manipulation

Algebraic manipulation refers to the techniques used to rearrange and simplify algebraic expressions. To find the value of C in the equation F(-1) = 4, we will substitute -1 into the function F(x) and solve for C. Mastery of algebraic manipulation is essential for solving equations and understanding calculus concepts.
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Constant Term in Functions

In the context of functions, a constant term is a fixed value that does not change with the input variable. In the function F(x) = x² - 3x + C, the term C represents a constant that shifts the graph of the function vertically. Understanding how constant terms affect function values is crucial for analyzing and interpreting functions in calculus.
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