Math Problem Statement

Find the stationary points and rough points and endpoint for f(x)=x^2+3-1x4\le x\le4

Solution

Let's solve this step-by-step. We are asked to find the stationary points, rough points (discontinuities or non-differentiable points), and endpoints of the function f(x)=x2+3f(x) = x^2 + 3 on the closed interval 1x4-1 \leq x \leq 4.


Step 1: Stationary Points

Stationary points occur when the first derivative of the function is zero.

  1. Compute the derivative of f(x)f(x): f(x)=ddx(x2+3)=2xf'(x) = \frac{d}{dx}(x^2 + 3) = 2x
  2. Set the derivative equal to zero to find the critical points: 2x=0x=02x = 0 \quad \Rightarrow \quad x = 0

Thus, the only stationary point within the interval is x=0x = 0.

  1. Evaluate the function at x=0x = 0: f(0)=02+3=3f(0) = 0^2 + 3 = 3

So, the stationary point is (0,3)(0, 3).


Step 2: Endpoints

Since the function is defined on a closed interval 1x4-1 \leq x \leq 4, we must evaluate it at the endpoints.

  • At x=1x = -1: f(1)=(1)2+3=4f(-1) = (-1)^2 + 3 = 4

  • At x=4x = 4: f(4)=42+3=16+3=19f(4) = 4^2 + 3 = 16 + 3 = 19

Thus, the function takes the values:

  • At x=1x = -1: f(1)=4f(-1) = 4
  • At x=4x = 4: f(4)=19f(4) = 19

Step 3: Non-differentiable Points (Rough Points)

  • The function f(x)=x2+3f(x) = x^2 + 3 is a polynomial, which is continuous and differentiable everywhere on R\mathbb{R}.
  • Thus, there are no rough points or non-differentiable points in the given interval 1x4-1 \leq x \leq 4.

Summary

  • Stationary point: (0,3)(0, 3)
  • Endpoints: (1,4)(-1, 4) and (4,19)(4, 19)
  • No rough points (the function is differentiable throughout the interval).

Optional: Graphing Insight

The function f(x)=x2+3f(x) = x^2 + 3 is a parabola that opens upwards.

  • The stationary point (0,3)(0, 3) is the global minimum in the given interval.
  • The values increase as we move towards the endpoints.

Do you want further clarifications or have any questions?


Here are five related questions to deepen understanding:

  1. What is the difference between a stationary point and an inflection point?
  2. How do we determine whether a stationary point is a minimum or maximum?
  3. What happens if the derivative is undefined at some point in the interval?
  4. How do we find global extrema for functions defined on closed intervals?
  5. How do endpoints affect the extrema in a given interval?

Tip: Always check both endpoints and stationary points when finding extrema on a closed interval.

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Math Problem Analysis

Mathematical Concepts

Calculus
Stationary Points
Differentiation
Continuity
Quadratic Functions

Formulas

Derivative: f'(x) = 2x
Stationary Points: Set derivative to zero, f'(x) = 0
Endpoints: Evaluate f(x) at the interval boundaries

Theorems

Differentiability of polynomials: Polynomials are differentiable everywhere
Extrema on closed intervals: The Extreme Value Theorem

Suitable Grade Level

Grades 10-12