Math Problem Statement

for f(x) = x^2 + 3x + 1 find f'(1) using f'(a)= lim x->a (f(x)-f(a))/(x-a)

Solution

To find f(1)f'(1) using the definition of the derivative, which is:

f(a)=limxaf(x)f(a)xaf'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}

In this case, a=1a = 1, so we need to compute f(1)f'(1). The function is f(x)=x2+3x+1f(x) = x^2 + 3x + 1, and we proceed with the following steps:

Step 1: Compute f(1)f(1)

We first evaluate f(1)f(1):

f(1)=(1)2+3(1)+1=1+3+1=5f(1) = (1)^2 + 3(1) + 1 = 1 + 3 + 1 = 5

Step 2: Write the difference quotient

Now, we use the difference quotient formula:

f(1)=limx1f(x)f(1)x1f'(1) = \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1}

Substitute f(x)=x2+3x+1f(x) = x^2 + 3x + 1 and f(1)=5f(1) = 5:

f(1)=limx1(x2+3x+1)5x1f'(1) = \lim_{x \to 1} \frac{(x^2 + 3x + 1) - 5}{x - 1}

Step 3: Simplify the expression

Simplify the numerator:

f(1)=limx1x2+3x+15x1=limx1x2+3x4x1f'(1) = \lim_{x \to 1} \frac{x^2 + 3x + 1 - 5}{x - 1} = \lim_{x \to 1} \frac{x^2 + 3x - 4}{x - 1}

Now, factor the quadratic x2+3x4x^2 + 3x - 4:

x2+3x4=(x1)(x+4)x^2 + 3x - 4 = (x - 1)(x + 4)

So the expression becomes:

f(1)=limx1(x1)(x+4)x1f'(1) = \lim_{x \to 1} \frac{(x - 1)(x + 4)}{x - 1}

Step 4: Cancel common terms

Cancel the (x1)(x - 1) terms:

f(1)=limx1(x+4)f'(1) = \lim_{x \to 1} (x + 4)

Step 5: Evaluate the limit

Now, evaluate the limit as x1x \to 1:

f(1)=1+4=5f'(1) = 1 + 4 = 5

Thus, f(1)=5f'(1) = 5.

Would you like further clarification, or details on any of the steps?

Here are five related questions to deepen your understanding:

  1. How would you compute f(a)f'(a) for any general value of aa using this same process?
  2. How does the result compare with using basic derivative rules like the power rule for this problem?
  3. What happens when a quadratic function is used in the limit definition of the derivative?
  4. How would you approach a similar problem if the function were cubic instead of quadratic?
  5. Can you think of a case where factoring might not work directly, and how could you handle it?

Tip: Always try simplifying the numerator in a difference quotient by factoring to cancel common terms with the denominator.

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Math Problem Analysis

Mathematical Concepts

Calculus
Limits
Derivatives
Quadratic Functions

Formulas

f'(a) = lim (x → a) (f(x) - f(a)) / (x - a)

Theorems

Limit definition of the derivative

Suitable Grade Level

Grades 11-12