Question

Shown above is the graph of the differentiable function f, along with the line tangent to the graph of f at x=2. What is the value of f′(2) ?

Answer 1

f'(2) means "the slope of the tangent line to y=f(x) when x=2".

Since that tangent line is drawn, this is really asking, "What's the slope of that line?"

So, what's the rise/run for that line?

Answer 2

The value of f'(2) is the slope of the tangent line, which is $\frac{1}{2}$, given by option a.

The tangent line to a function f(x) at a point $x_0$ is given by, in point-slope form:

$y - f(x_0) = f^{\prime}(x_0)(x - x_0)$

That is, the slope of the line is the derivative of the function at $x = x_0$.

• Given two points, the slope is given by change in y divided by change in x.

• For the tangent line, there are points (0,2) and (2,3).

• Thus, the slope is:

$f^{\prime}(2) = \frac{3 - 2}{2 - 0} = \frac{1}{2}$

Given by option a.

A similar problem is given at https://brainly.com/question/22426360