Math Problem Statement

use this graph to find the following derivative rules. a. f'(-2.5)= b. f'(0)= c. f'(1)= d. f'(4)= e. f'(6)=

Use this graph to find the following derivative rules. a. f'(-2.5)= b. f'(0)= c. f'(1)= d. f'(4)= e. f'(6)=

Solution

Let's analyze the piecewise linear graph you provided to determine the derivative values at the specific points.

General Approach for Derivatives of Linear Segments:

  • Since the graph consists of straight line segments, the derivative (slope) at each point where the function is smooth is constant.
  • At points where the graph has a sharp turn (like at x=0x = 0 and x=1x = 1), the derivative is undefined, as the graph is not smooth at those points.

Let's answer the derivative rules based on the graph:

a. f(2.5)f'(-2.5):

  • From x=8x = -8 to x=0x = 0, the graph is a straight line increasing. We can calculate the slope between two points. The slope of this line is calculated by Δy/Δx\Delta y / \Delta x. For example, from (8,1)(-8, -1) to (0,1)(0, 1), the slope is: slope=1(1)0(8)=28=0.25\text{slope} = \frac{1 - (-1)}{0 - (-8)} = \frac{2}{8} = 0.25 Therefore, f(2.5)=0.25f'(-2.5) = 0.25.

b. f(0)f'(0):

  • At x=0x = 0, there is a sharp corner (cusp), indicating a non-differentiable point. Therefore, f(0)f'(0) is undefined.

c. f(1)f'(1):

  • At x=1x = 1, there is another sharp turn in the graph, indicating non-differentiability. Therefore, f(1)f'(1) is undefined.

d. f(4)f'(4):

  • From x=1x = 1 to x=4x = 4, the graph is decreasing. The slope can be calculated between (1,0)(1, 0) and (4,2)(4, -2): slope=2041=23=23\text{slope} = \frac{-2 - 0}{4 - 1} = \frac{-2}{3} = -\frac{2}{3} Therefore, f(4)=23f'(4) = -\frac{2}{3}.

e. f(6)f'(6):

  • From x=4x = 4 to x=8x = 8, the graph is sharply increasing. The slope can be calculated between (4,2)(4, -2) and (8,8)(8, 8): slope=8(2)84=104=2.5\text{slope} = \frac{8 - (-2)}{8 - 4} = \frac{10}{4} = 2.5 Therefore, f(6)=2.5f'(6) = 2.5.

Summary of Derivatives:

  • f(2.5)=0.25f'(-2.5) = 0.25
  • f(0)f'(0) is undefined
  • f(1)f'(1) is undefined
  • f(4)=23f'(4) = -\frac{2}{3}
  • f(6)=2.5f'(6) = 2.5

Would you like further explanation on any of these calculations or details on slopes?

Related Questions:

  1. What is the derivative at x=5x = -5?
  2. How does differentiability change at points where the graph has sharp turns?
  3. What is the physical meaning of a positive versus negative derivative?
  4. How would you interpret a zero derivative from a graph like this?
  5. Can you describe the general shape of the graph based on its slopes?

Tip: The derivative at a point provides information about the slope or rate of change at that point. If the graph has a sharp corner, the derivative is not defined there.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivative
Slope
Differentiability

Formulas

Slope formula: m = (y2 - y1) / (x2 - x1)
Derivative is the slope of the tangent line to the curve at a point

Theorems

Non-differentiability at sharp turns or cusps

Suitable Grade Level

Grades 11-12