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Chapter 1: Problem 51

Consider the function \(k\) given by $$k(x)=|x-3|+2$$ a) For what \(x\) -value(s) is the function not differentiable? b) Evaluate \(k^{\prime}(0), k^{\prime}(1), k^{\prime}(4),\) and \(k^{\prime}(10)\).

Short Answer

Expert verified
Function is not differentiable at \(x=3\). Derivatives: \(-1, -1, 1, 1\).

Step by step solution

01

Understanding Non-differentiability of Absolute Value Function

The absolute value function \(|x-a|\) is not differentiable at \(x = a\). Here, the function \(k(x) = |x-3| + 2\) involves an absolute value term where \(a = 3\). Thus, the function \(k(x)\) is not differentiable at \(x = 3\) because the absolute value function has a sharp corner at this point.
02

Calculating Derivatives for Different Intervals

To evaluate the derivatives at given points, first consider the piecewise nature of \(k(x) = |x-3| + 2\). We split it into two cases: 1. If \(x \geq 3\), then \(|x-3| = x-3\), so \(k(x) = x-1\), and its derivative \(k'(x) = 1\).2. If \(x < 3\), then \(|x-3| = 3-x\), so \(k(x) = 5-x\), and its derivative \(k'(x) = -1\).
03

Evaluating Derivatives at Specific Points

Evaluate the derivative \(k'(x)\) at the points given:- For \(x = 0\): Since \(0 < 3\), use \(k'(x) = -1\). So, \(k'(0) = -1\).- For \(x = 1\): Since \(1 < 3\), use \(k'(x) = -1\). So, \(k'(1) = -1\).- For \(x = 4\): Since \(4 \geq 3\), use \(k'(x) = 1\). So, \(k'(4) = 1\).- For \(x = 10\): Since \(10 \geq 3\), use \(k'(x) = 1\). So, \(k'(10) = 1\).

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Key Concepts

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

Piecewise Functions
Piecewise functions are a fascinating part of calculus. They are defined by different expressions depending on the input value. These functions have multiple sub-functions, each with its own interval of validity. This is exactly what happens with the function \(k(x) = |x-3| + 2\), which can be rewritten into two separate linear functions based on the value of \(x\):
  • If \(x \geq 3\), \(k(x)\) aligns with the line \(k(x) = x - 1\).
  • If \(x < 3\), \(k(x)\) follows \(k(x) = 5 - x\).
Such piecewise definitions allow for flexibility in modeling different real-world situations, where a single function doesn't fit all scenarios. Understanding how to approach these different pieces helps in simplifying calculus problems.
Absolute Value Functions
Absolute value functions are unique because they transform all negative inputs to positive values. Graphically, these functions create a characteristic "V" shape with a vertex at the point where the expression inside the absolute value is zero. For \(k(x) = |x-3| + 2\), the absolute value \(|x-3|\) becomes zero at \(x=3\).
Absolute value functions are rewritten as piecewise functions to manage the operations within different intervals:
  • If the expression inside is non-negative, such as \(x-3\) for \(x\geq 3\), the absolute value does not alter it.
  • If the expression inside is negative, as in \(3-x\) for \(x < 3\), it negates the expression to ensure positivity.
Learning to handle these transformations is crucial for solving more complex calculus problems that involve absolute values.
Non-differentiability
Differentiability is an essential concept in calculus, indicating whether a function has a derivative at a given point. Some functions, like absolute value functions, are not differentiable everywhere.
For the function \(k(x) = |x-3| + 2\), non-differentiability happens at \(x=3\). Why? Because there's a sharp corner or cusp at this point due to the absolute value function. This corner prevents a tangent line from being drawn, resulting in the function lacking a derivative there.
Recognizing non-differentiability is vital for accurate derivative evaluation and understanding a function's behavior.
Derivative Evaluation
Evaluating the derivatives of piecewise functions involves examining each segment separately. For the piecewise representation of our original function \(k(x)\):
  • For \(x < 3\), the expression is \(5-x\) with a derivative \(k'(x) = -1\).
  • For \(x \geq 3\), the expression is \(x-1\) with a derivative \(k'(x) = 1\).
Given specific \(x\) values, we verify which piece of the function applies and compute the derivative accordingly. This involves checking each value to decide the function's form:
  • At \(x=0\) and \(x=1\), both less than 3, the derivative is \(-1\).
  • At \(x=4\) and \(x=10\), both greater or equal to 3, the derivative is \(1\).
Each evaluated derivative helps graph and understand how each section of the function behaves, leading to a complete picture of the function's dynamics.

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Most popular questions from this chapter

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=-x^{3}+x^{2}+5 x-1 $$

Graph fand \(f^{\prime}\) and then determine \(f^{\prime}(1) .\) $$ f(x)=\frac{5 x^{2}+8 x-3}{3 x^{2}+2} $$

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} $$

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=\frac{1}{3} x^{3}-3 x+2 $$

Graph the function \(f\) given by $$ f(x)=\left\\{\begin{array}{ll} -3, & \text { for } x=-2 \\ x^{2}, & \text { for } x \neq-2 \end{array}\right. $$ Use GRAPH and TRACE to find each of the following limits. When necessary, state that the limit does not exist. a) \(\lim _{x \rightarrow-2^{+}} f(x)\) b) \(\lim _{x \rightarrow-2^{-}} f(x)\) c) \(\lim _{x \rightarrow-2} f(x)\) d) \(\lim _{x \rightarrow 2^{+}} f(x)\) e) \(\lim _{x \rightarrow 2^{-}} f(x)\) f) Does \(\lim _{x \rightarrow-2} f(x)=f(-2)\) ? g) Does \(\lim _{x \rightarrow 2} f(x)=f(2)\) ?
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