Math Problem Statement

Use the limit definition to compute the derivative of the function f(x) = √x + |2x| at x = 1. Find an equation of the tangent line to the graph of f at x = 1.

Solution

The image presents two tasks involving derivatives and tangent lines:

  1. Task 1: Use the limit definition to compute the derivative of the function f(x)=x+∣2x∣f(x) = \sqrt{x} + |2x| at x=1x = 1.

  2. Task 2: Find the equation of the tangent line to the graph of ff at x=1x = 1.

Let's break this down and solve each task.

Task 1: Compute the Derivative Using the Limit Definition

The limit definition of the derivative is:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Given f(x)=x+∣2x∣f(x) = \sqrt{x} + |2x|, we compute f′(1)f'(1).

Step 1: Evaluate f(1)f(1):

f(1)=1+∣2(1)∣=1+2=3f(1) = \sqrt{1} + |2(1)| = 1 + 2 = 3

Step 2: Use the limit definition at x=1x = 1:

f′(1)=lim⁡h→0f(1+h)−f(1)hf'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h}

Substitute f(x)=x+∣2x∣f(x) = \sqrt{x} + |2x| into this expression:

f(1+h)=1+h+∣2(1+h)∣=1+h+2(1+h)f(1+h) = \sqrt{1+h} + |2(1+h)| = \sqrt{1+h} + 2(1+h)

Now, substitute f(1+h)f(1+h) and f(1)=3f(1) = 3:

f′(1)=lim⁡h→0(1+h+2(1+h))−3hf'(1) = \lim_{h \to 0} \frac{\left(\sqrt{1+h} + 2(1+h)\right) - 3}{h} Simplify the numerator:

= \lim_{h \to 0} \frac{\sqrt{1+h} - 1 + 2h}{h}$$ We can split the limit: $$f'(1) = \lim_{h \to 0} \frac{\sqrt{1+h} - 1}{h} + \lim_{h \to 0} 2$$ The second limit is simply 2. For the first limit, we use the approximation of the square root: $$\frac{\sqrt{1+h} - 1}{h} \approx \frac{h}{2h} = \frac{1}{2}$$ Thus: $$f'(1) = \frac{1}{2} + 2 = \frac{5}{2}$$ ### **Task 2: Equation of the Tangent Line** The equation of the tangent line to $$f(x)$$ at $$x = 1$$ is given by: $$y - f(1) = f'(1)(x - 1)$$ We already know that $$f'(1) = \frac{5}{2}$$ and $$f(1) = 3$$. Substituting these values into the equation: $$y - 3 = \frac{5}{2}(x - 1)$$ Simplify: $$y = \frac{5}{2}(x - 1) + 3$$ $$y = \frac{5}{2}x - \frac{5}{2} + 3$$ $$y = \frac{5}{2}x + \frac{1}{2}$$ Thus, the equation of the tangent line is: $$y = \frac{5}{2}x + \frac{1}{2}$$ --- Would you like further details or have any questions? ### 5 Related Questions: 1. How do you apply the limit definition to more complex functions? 2. What are common strategies to simplify limits involving square roots? 3. How is the absolute value function treated when differentiating? 4. Can you use an alternative method, such as differentiation rules, to find $$f'(x)$$? 5. How does the slope of the tangent line relate to the derivative? ### Tip: When dealing with square roots in limits, multiplying by the conjugate can help simplify the expression.

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Math Problem Analysis

Mathematical Concepts

Calculus
Limit Definition of Derivative
Tangent Line
Absolute Value
Square Root Function

Formulas

Limit definition of the derivative: f'(x) = lim_{h -> 0} (f(x+h) - f(x)) / h
Tangent line equation: y - f(a) = f'(a)(x - a)

Theorems

Limit Definition of Derivative

Suitable Grade Level

Grades 11-12

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