If a tangent to the graph of a function $y = f(x)$ at a point with abscissa $x_0$ forms an angle of 45° with the positive direction of the x-axis, what is the value of $f'(x_0)$?
Understand the Problem
The question is asking about the relationship between the angle a tangent line makes with the x-axis and the derivative of the function at that point. Specifically, it states that the tangent line to y = f(x) at x = x_0 forms a 45-degree angle with the positive x-axis. We must calculate the value of f'(x_0).
Answer
$f'(x_0) = 1$
Answer for screen readers
$f'(x_0) = 1$
Steps to Solve
- Relate the angle to the slope
The tangent of the angle that a line makes with the positive x-axis is equal to the slope of the line. Let the angle be $\theta$. Then the slope $m$ is given by:
$m = \tan(\theta)$
- Calculate the slope
We are given that the angle is 45 degrees. Therefore, $\theta = 45^\circ$. We need to find the tangent of this angle:
$m = \tan(45^\circ)$
Since $\tan(45^\circ) = 1$, the slope of the tangent line is 1.
- Relate the slope to the derivative
The derivative of a function $f(x)$ at a point $x = x_0$, denoted as $f'(x_0)$, gives the slope of the tangent line to the graph of $f(x)$ at that point. Therefore:
$f'(x_0) = m$
- Find the value of $f'(x_0)$
Since we know the slope $m = 1$, we can conclude that:
$f'(x_0) = 1$
$f'(x_0) = 1$
More Information
The derivative of a function at a given point is a fundamental concept in calculus. It represents the instantaneous rate of change of the function at that point, which geometrically corresponds to the slope of the tangent line to the function's graph at that point.
Tips
A common mistake would be to compute $\tan(45^\circ)$ incorrectly, or to not remember that the tangent of the angle is the slope. Also, one might confuse the angle with the slope directly without using the tangent function.
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