Answer:
Step-by-step explanation:
x\sin\left(\frac{1}{x}\right), & x \\eq 0 \\
0, & x = 0
\end{cases}\]
has a tangent at the origin, which is \(x = 0\).
To find the tangent at \(x = 0\), we need to calculate the derivative of \(f(x)\) at that point. We will use the limit definition of the derivative:
\[f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h}\]
1. **Find \(f(0)\):
Since \(f(0) = 0\) (as defined in the piecewise function), we have \(f(0) = 0\).
2. **Find \(f(h)\) for \(h \\eq 0\)**:
For \(x \\eq 0\), \(f(x) = x\sin\left(\frac{1}{x}\right)\). So, when \(x = h\), we have \(f(h) = h\sin\left(\frac{1}{h}\right)\).
3. **Calculate the difference quotient**:
Now, let's plug these values into the difference quotient formula:
\[f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h\sin\left(\frac{1}{h}\right) - 0}{h}\]
4. **Simplify the expression**:
We can simplify this expression further:
\[f'(0) = \lim_{h \to 0} \sin\left(\frac{1}{h}\right)\]
5. **Analyze the limit as \(h\) approaches 0**:
The function \(\sin\left(\frac{1}{h}\right)\) oscillates wildly as \(h\) approaches 0. As \(h\) gets closer to 0, the sine function takes on values between -1 and 1 infinitely many times, and there is no single value that the limit converges to. Therefore, the limit does not exist.
Since the limit of the difference quotient does not exist, there is no unique tangent at the origin (\(x = 0\)). The graph of \(f(x)\) does not have a tangent at this point because the derivative is not well-defined at \(x = 0\).
In summary, there is no tangent line to the graph of \(f(x)\) at the origin.