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Chapter 3: Problem 44

If \(f:[-2 a, 2 a] \rightarrow R\) is an odd function such that \(f(x)=f(2 a-x)\) for \(x \in(a, 2 a)\). if the left hand derivative of \(f(x)\) at \(x=a\) is zero, then the left hand derivative of \(f(x)\) at \(x=-a\) is (A) 1 (B) \(-1\) (C) 0 (T)) none

Short Answer

Expert verified
Answer: The left-hand derivative of \(f(x)\) at \(x=-a\) is 0.

Step by step solution

01

Recall properties of odd function

An odd function is a function that satisfies the property \(f(-x) = -f(x)\) for all values of \(x\) in its domain.
02

Evaluate the given function for the specific domain

As the function is defined for \(x \in (a, 2a)\), we can find the relationship between \(f(a-x)\) and \(f(x)\) using the given condition \(f(x) = f(2a-x)\). Let's first replace \(x\) with \(a-x\): \(f(a - x) = f(2a - (a-x))\) This simplifies to: \(f(a - x) = f(a + x)\) Now we know the relationship between \(f(a - x)\) and \(f(a+x)\) for the given domain.
03

Apply the definition of left-hand derivative at \(x=-a\)

To find the left-hand derivative of \(f(x)\) at \(x=-a\) we need to evaluate: \(\lim_{h \to 0^{-}} \frac{f(-a + h) - f(-a)}{h}\) Using the relationship from Step 2: \(f(a - x) = f(a + x)\), we can write \(f(-a + h)\) as \(f(a - h)\), thus our left-hand derivative becomes: \(\lim_{h \to 0^{-}} \frac{f(a - h) - f(-a)}{h}\)
04

Use the properties of odd function to find the left-hand derivative at \(x=-a\)

Since \(f(x)\) is an odd function, we can replace \(f(-a)\) with \(-f(a)\). Now our limit is: \(\lim_{h \to 0^{-}} \frac{f(a - h) + f(a)}{h}\) As the left-hand derivative of \(f(x)\) at \(x=a\) is zero, we have: \(\lim_{h \to 0^{-}} \frac{f(a - h) - f(a)}{-h} = 0\) Now, multiplying both the numerator and the denominator of our limit with \(-1\), we get: \(\lim_{h \to 0^{-}} \frac{f(a) - f(a - h)}{h} = 0\) Comparing with the left-hand derivative at \(x=-a\), we find that they are equal: \(\lim_{h \to 0^{-}} \frac{f(a - h) + f(a)}{h} = \lim_{h \to 0^{-}} \frac{f(a) - f(a - h)}{h} = 0\) Hence, the left-hand derivative of \(f(x)\) at \(x=-a\) is 0 which is option (C).

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Most popular questions from this chapter

The function $f(x)=\frac{|x|-x\left(3^{1 / x}+1\right)}{3^{1 / x}-1}, x \neq 0, f(0)=0$ is (A) discontinuous at \(x=0\) (B) continuous at \(\mathrm{x}=0\) but not differentiable there (C) both continuous and differentiable at \(\mathrm{x}=0\) (D) differentiable but not continuous at \(\mathrm{x}=0\)

Let $\mathrm{f}(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{\left(\mathrm{x}^{2}+2 \mathrm{x}+3+\sin \pi \mathrm{x}\right)^{\mathrm{n}}-1}{\left(\mathrm{x}^{2}+2 \mathrm{x}+3+\sin \pi \mathrm{x}\right)^{\mathrm{n}}+1}$, then (A) \(\mathrm{f}(\mathrm{x})\) is continuous and differentiable for all \(\mathrm{x} \in \mathrm{R}\). (B) \(f(x)\) is continuous but not differentiable for all \(x \in R\). (C) \(\mathrm{f}(\mathrm{x})\) is discontinuous at infinite number of points. (D) \(\mathrm{f}(\mathrm{x})\) is discontinuous at finite number of points.

Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) satisfying $|\mathrm{f}(\mathrm{x})| \leq \mathrm{x}^{2} \forall \mathrm{x} \in \mathrm{R}$, then (A) ' \(f\) is continuous but non-differentiable at \(x=0\) (B) ' \(\mathrm{f}^{\prime}\) is discontinuous at \(\mathrm{x}=0\) (C) ' \(\mathrm{f}^{\prime}\) is differentiable at \(\mathrm{x}=0\) (D) None of these

If $f(x)=\left\\{\begin{array}{ll}{[x]+\sqrt{\\{x\\}}} & x<1 \\\ \frac{1}{[x]+\\{x\\}^{2}} & x \geq 1\end{array}\right.$, then [where [. ] and \\{ - \(\\}\) represent greatest integer and fractional part functions respectively] (A) \(\mathrm{f}(\mathrm{x})=\) is continuous at \(\mathrm{x}=1\) but not differentiable (B) \(\mathrm{f}(\mathrm{x})\) is not continuous at \(\mathrm{x}=1\)

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