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Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). $$ f(x)=x^{3}+2 x^{2}+1 $$

Short Answer

Expert verified
The derivative is \( 3x^2 + 4x \).

Step by step solution

01

Substitute into the Difference Quotient

The first step is to substitute the given function \( f(x) = x^3 + 2x^2 + 1 \) into the definition of the derivative: \[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \] So we need to compute \( f(x+h) \).
02

Calculate \( f(x+h) \)

We substitute \( x+h \) into the function \( f(x) \):\[ f(x+h) = (x+h)^3 + 2(x+h)^2 + 1 \]Expanding this, we get:\[ f(x+h) = (x^3 + 3x^2 h + 3xh^2 + h^3) + 2(x^2 + 2xh + h^2) + 1 \] This simplifies to:\[ f(x+h) = x^3 + 3x^2 h + 3xh^2 + h^3 + 2x^2 + 4xh + 2h^2 + 1 \],and further simplifies to\[ = x^3 + 2x^2 + 1 + (3x^2 + 4x)h + (3x + 2)h^2 + h^3\].
03

Substitute \( f(x+h) \) and \( f(x) \) back

Now, plug \( f(x+h) \) and \( f(x) \) back into the difference quotient:\[ f'(x) = \lim_{{h \to 0}} \frac{{[x^3 + 2x^2 + 1 + (3x^2 + 4x)h + (3x + 2)h^2 + h^3] - [x^3 + 2x^2 + 1]}}{h} \].The \( x^3 + 2x^2 + 1 \) terms cancel out, leaving:\[ f'(x) = \lim_{{h \to 0}} \frac{{(3x^2 + 4x)h + (3x + 2)h^2 + h^3}}{h} \].
04

Simplify the Fraction

Factor out \( h \) from the numerator:\[ f'(x) = \lim_{{h \to 0}} \frac{h(3x^2 + 4x + (3x + 2)h + h^2)}{h} \].Cancel \( h \) from the numerator and denominator:\[ f'(x) = \lim_{{h \to 0}} (3x^2 + 4x + (3x + 2)h + h^2) \].
05

Evaluate the Limit

Evaluate the limit as \( h \to 0 \):\[ f'(x) = 3x^2 + 4x \].The \( (3x + 2)h + h^2 \) terms vanish as \( h \to 0 \).
06

Conclusion on Derivative

The derivative of the function \( f(x) = x^3 + 2x^2 + 1 \) with respect to \( x \) is \( 3x^2 + 4x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference Quotient
The difference quotient is a fundamental concept in calculus, used to calculate the derivative of a function. It is represented by the formula \( \frac{f(x+h) - f(x)}{h} \). This expression estimates the rate at which the function \( f(x) \) changes at a point \( x \). By taking the value of the function at a small increment \( h \) and subtracting the original function value, we focus on the change over a tiny interval.
The goal is to determine how \( f(x) \) behaves as \( h \) becomes very small.
  • The numerator of the difference quotient gives the change in the function value.
  • The denominator, \( h \), represents the small interval over which we measure this change.

Calculating the difference quotient is the preliminary step in finding a derivative, which serves as the rate of change or the slope of the tangent line at a given point.
Limit Process
The limit process is essential for making the difference quotient a powerful tool in calculus. By letting \( h \to 0 \), we make sure that our estimate of the rate of change becomes more accurate. The more \( h \) approaches zero, the closer we get to finding the true instantaneous rate of change or the derivative at a point. In mathematical terms, this is expressed as:\[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]
The importance of the limit lies in transforming an average rate of change over an interval into an exact rate of change at a single point.
  • The limit ensures precision by reducing \( h \) to an infinitesimally small value.
  • This process helps in understanding smooth curves and steepness.
As \( h \) becomes negligible, the last two terms that contain \( h \) in \( (3x + 2)h + h^2 \) disappear, just leaving us with the exact derivative value.
Polynomial Functions
Polynomial functions, such as \( f(x) = x^3 + 2x^2 + 1 \), are expressions involving variables raised to whole-number exponents and coefficients. These functions are crucial because their derivatives are easier to compute, given their smooth and continuous nature.
To find derivatives of polynomial functions:
  • Each term is treated separately.
  • The power rule is frequently applied.

For our exercise, we have three terms; each needs individual attention before piecing the overall function derivative together. Polynomial functions provide a straightforward way for practicing differentiation due to their predictable pattern and ease in manipulation.
Differentiation Techniques
Differentiation techniques are methods used to find the derivative of a function. For polynomial functions like \( f(x) = x^3 + 2x^2 + 1 \), these techniques involve well-defined rules.
In our exercise:
  • Use the power rule for each term: If \( f(x) = ax^n \), then \( f'(x) = anx^{n-1} \).
  • Apply linearity: Differentiate polynomials by handling each term individually, then combine.

When using the power rule:
  • The term \( x^3 \) becomes \( 3x^2 \).
  • The term \( 2x^2 \) turns into \( 4x \).
  • Constant terms, like \( 1 \), vanish as their derivative is zero.

Implementing these techniques shows that the derivative of \( f(x) = x^3 + 2x^2 + 1 \) is \( 3x^2 + 4x \). This powerful simplification comes through applying differentiation techniques skillfully.

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