Chapter 2: Problem 86
Determining Differentiability In Exercises \(85-88\) , find the derivatives from the left and from the right at \(x=1\) (if they exist). Is the function differentiable at \(x=1 ?\) $$ f(x)=\sqrt{1-x^{2}} $$
Short Answer
Expert verified
The function \(f(x)=\sqrt{1-x^2}\) is not differentiable at \(x=1\), as neither the left-hand nor the right-hand derivatives at this point exist.
Step by step solution
01
Establish the Function and Point of Interest
We have the function \(f(x)=\sqrt{1-x^2}\) and we want to find if it is differentiable or not at \(x=1\). The differentiability at a point depends on the left-hand and right-hand derivative at that point.
02
Compute Left-hand Derivative
The left-hand derivative, denoted as \(f'_-(1)\), is computed through the limit as \(h\) approach zero from negative side, i.e., \[\lim_{{h \to 0^{-}}} \frac{f(1+h)-f(1)}{h}\]. Substituting for \(f(x)\), we get \[\lim_{{h \to 0^-}} \frac{\sqrt{1-(1+h)^2}-\sqrt{1-1^2}}{h}\]. When simplified, due to the square root of a negative number, this limit does not exist.
03
Compute Right-hand Derivative
The right-hand derivative, denoted \(f'_+(1)\), is computed similarly, through the limit as \(h\) approach zero from positive side, i.e., \[\lim_{{h \to 0^{+}}} \frac{f(1+h)-f(1)}{h}\]. Substituting for \(f(x)\), we get \[\lim_{{h \to 0^+}} \frac{\sqrt{1-(1+h)^2}-\sqrt{1-1^2}}{h}\]. Similarly, due to the square root of a negative number, this limit also does not exist.
04
Determine Differentiability
A function is differentiable at a point if the left-hand derivative and right-hand derivative at that point exist and are equal. Here, neither derivative exists, hence the function \(f(x)=\sqrt{1-x^2}\) is not differentiable at \(x=1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Left-hand Derivative
The left-hand derivative of a function at a specific point captures the instantaneous rate of change of the function as the input approaches the given point from the left. In mathematical terms, if we're interested in the left-hand derivative of a function at a point, say, at x=1 , we would look at the limit of the difference quotient as h approaches zero from the negative side:
\[ f'_{-}(1) = \blim_{{h \to 0^{-}}} \frac{f(1 + h) - f(1)}{h} \]
In simpler terms, think ofh as a tiny step to the left of our point of interest x=1 . As the step becomes infinitesimally small, we get closer to the actual slope of the function at exactly x=1 , approaching from the left. Consider viewing this as zooming in on the graph at x=1 until the curve looks almost like a straight line, and then measuring the slope of that line.
\[ f'_{-}(1) = \blim_{{h \to 0^{-}}} \frac{f(1 + h) - f(1)}{h} \]
In simpler terms, think of
Right-hand Derivative
Conversely, the right-hand derivative refers to the instantaneous rate of change of the function as the input approaches the point from the right. The right-hand derivative at a point x=1 is mathematically represented by the limit of the difference quotient as h approaches zero from the positive side:
\[ f'_{+}(1) = \blim_{{h \to 0^{+}}} \frac{f(1 + h) - f(1)}{h} \]
Again, ifh is a minuscule step, this time to the right of our point of interest, as it shrinks to zero, we approach the exact slope at x=1 , but from the right. A helpful visualization is to imagine a tiny ant walking towards x=1 from the right side, and just as it's about to reach x=1 , we calculate the slope of the path it has traveled. In the exercise provided, neither the left-hand nor the right-hand limit exists due to the square root of a negative number that arises, highlighting the importance of these limits in evaluating differentiability.
\[ f'_{+}(1) = \blim_{{h \to 0^{+}}} \frac{f(1 + h) - f(1)}{h} \]
Again, if
Limit of a Function
The limit of a function encapsulates the value that a function approaches as its input approaches some value. It is crucial in calculus because it helps define both the continuity and differentiability of functions. When we say \( \blim_{{x \to c}} f(x) = L \), we mean that as x gets very close to c , the function f(x) gets arbitrarily close to L . However, the function doesn't necessarily have to reach L when x=c .
In terms of differentiability, if we cannot define a limit as we approach from either the left or the right side of a point, as with the square root function in our exercise, the concept indicates that the function may have a corner, cusp, or discontinuity at that point. In these cases, the function is not differentiable there, as there's no singular, well-defined slope.
In terms of differentiability, if we cannot define a limit as we approach from either the left or the right side of a point, as with the square root function in our exercise, the concept indicates that the function may have a corner, cusp, or discontinuity at that point. In these cases, the function is not differentiable there, as there's no singular, well-defined slope.
Continuity and Differentiability
Continuity and differentiability are two foundational concepts in calculus that are intimately connected. A function is continuous at a point if you can draw through that point without lifting your pencil. Formally, this means a function f(x) is continuous at a point x=c if \( \blim_{{x \to c}} f(x) = f(c) \). Differentiability, on the other hand, implies that a function has a well-defined tangent or rate of change at that point, which requires the function to be smooth or have no sharp corners or edges.
For a function to be differentiable at a point, it must be continuous there, but the converse isn't necessarily true — a continuous function might still have sharp turns that prevent differentiability. The exercise shows that the lack of left-hand and right-hand derivatives atx=1 results in the function f(x) = \sqrt{1 - x^{2}} being non-differentiable at that point, illustrating that continuity alone is not enough to guarantee differentiability.
For a function to be differentiable at a point, it must be continuous there, but the converse isn't necessarily true — a continuous function might still have sharp turns that prevent differentiability. The exercise shows that the lack of left-hand and right-hand derivatives at