shown above is the graph of a differentiable function F along with the line tangent to the graph of F at X=3. what is the value of f’3

Answer:
Option (A)
Step-by-step explanation:
Function 'f' given in the graph is differentiable with a tangent at x = 3.
Since, slope of the function = value of the differentiated function at a point
Tangent at a point x = 3 is passing through two points (3, 4) and (1, 5).
Slope of the function passing through [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is,
Slope = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
= [tex]\frac{5-4}{1-3}[/tex]
= [tex]-\frac{1}{2}[/tex]
Therefore, value of f'(3) will be equal to [tex]-\frac{1}{2}[/tex]
Option (A) will be the answer.
Tangent lines are lines that touches a curve at a point, without running over the curve.
The value of f'(3) is -1/2
From the figure, the line tangent to the graph passes through (5,3) and (1,5)
Next, we calculate the slope of the line using:
[tex]\mathbf{m = \frac{y_2 - y_1}{x_2 - x_1}}[/tex]
So, we have:
[tex]\mathbf{m = \frac{5-3}{1-5}}[/tex]
[tex]\mathbf{m = \frac{2}{-4}}[/tex]
Simplify
[tex]\mathbf{m = \frac{1}{-2}}[/tex]
Rewrite as:
[tex]\mathbf{m = -\frac{1}{2}}[/tex]
The line also passes through (3,4) and (1,5)
So, the slope of the line is:
[tex]\mathbf{m = \frac{y_2 - y_1}{x_2 - x_1}}[/tex]
So, we have:
[tex]\mathbf{m = \frac{5-4}{1-3}}[/tex]
[tex]\mathbf{m = \frac{1}{-2}}[/tex]
Rewrite as:
[tex]\mathbf{m = -\frac{1}{2}}[/tex]
The calculated slope is -1/2
This means that:
[tex]\mathbf{f'(3) = m}[/tex]
Substitute [tex]\mathbf{m = -\frac{1}{2}}[/tex]
[tex]\mathbf{f'(3) = -\frac{1}{2}}[/tex]
Hence, the value of f'(3) is -1/2
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