Respuesta :

mathmate
As given,
"tangent line to f(x) is tangent at (7,5)",  therefore f(7)=5.

Line is tangent at (7,5) and passes through (0,4), therefore the slope of the line is the slope of f(x=7) = f'(7).
[tex] f'(7) = slope\, of\, line = \frac{y_2-y_1}{x_2-x_1} = \frac{5-4}{7-0}=\frac{1}{7}[/tex]

frika

The equation of tangent line to y=f(x) at point [tex](x_0,y_0)[/tex] is

[tex]y=f'(x_0)(x-x_0)+y_0.[/tex]

In your case the tangent point is (7,5), then [tex]x_0=7,\ y_0=f(x_0)=f(7)=5[/tex] and the equation of tangent line is

[tex]y=f'(7)(x-7)+5.[/tex]

This tangent line passes through the point (0,4), then the coordinates of this point satisfy the equation:

[tex]4=f'(7)(0-7)+5,\\ \\f'(7)\cdot (-7)=4-5,\\ \\-7f'(7)=-1,\\ \\f'(7)=\dfrac{1}{7}.[/tex]

Answer: [tex]f(7)=5,\ f'(7)=\dfrac{1}{7}.[/tex]

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