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

shown above is the graph of a differentiable function F along with the line tangent to the graph of F at X3 what is the value of f3 class=

Respuesta :

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

Read more about tangent lines at:

https://brainly.com/question/23265136

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