Simplify the difference quotient ƒ(x+h)-ƒ(x)/h
ƒ(x) = 2x² -3x +1

Guida verificata passo dopo passo

Step 1: Start by substituting the function $ f(x) = 2x^2 - 3x + 1 $ into the difference quotient $ \frac{f(x+h) - f(x)}{h} $.

Step 2: Calculate $ f(x+h) $ by replacing $ x $ with $ x+h $ in the function: $ f(x+h) = 2(x+h)^2 - 3(x+h) + 1 $.

Step 3: Expand $ f(x+h) $: $ 2(x+h)^2 = 2(x^2 + 2xh + h^2) = 2x^2 + 4xh + 2h^2 $ and $ -3(x+h) = -3x - 3h $. Combine these to get $ f(x+h) = 2x^2 + 4xh + 2h^2 - 3x - 3h + 1 $.

Step 4: Substitute $ f(x+h) $ and $ f(x) $ into the difference quotient: $ \frac{(2x^2 + 4xh + 2h^2 - 3x - 3h + 1) - (2x^2 - 3x + 1)}{h} $.

Step 5: Simplify the expression by canceling out like terms and dividing each term by $ h $. This will give you the simplified form of the difference quotient.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Difference Quotient

The difference quotient is a formula used to find the average rate of change of a function over an interval. It is expressed as (ƒ(x+h) - ƒ(x))/h, where h represents a small change in x. This concept is fundamental in calculus as it leads to the definition of the derivative, which measures the instantaneous rate of change.

Video consigliato:

06:43

The Quotient Rule

Function Evaluation

Function evaluation involves substituting a specific value into a function to determine its output. In this case, we need to evaluate ƒ(x+h) and ƒ(x) for the given function ƒ(x) = 2x² - 3x + 1. Understanding how to correctly substitute and simplify expressions is crucial for manipulating the difference quotient.

Video consigliato:

Percorso guidato

4:26

Evaluating Composed Functions

Algebraic Simplification

Algebraic simplification is the process of reducing expressions to their simplest form. This involves combining like terms, factoring, and canceling common factors. In the context of the difference quotient, simplifying the expression after substituting ƒ(x+h) and ƒ(x) is essential to arrive at a clear and concise result, especially as h approaches zero.

Video consigliato: