a. To find the equation of the secant line through the points where \(x=5\) and \(x=8\), first calculate the function values at these points. Then, use the slope formula \((y_2 - y_1) / (x_2 - x_1)\) to find the slope of the secant line. Finally, use the point-slope form of a line to write the equation of the secant line.
b. To find the equation of the tangent line to the curve at \(x=5\), first find the derivative of the function \(f(x)\) to get the slope of the tangent line at any point \(x\). Evaluate this derivative at \(x=5\) to get the slope at that point. Then, use the point-slope form of a line with the point \((5, f(5))\) to write the equation of the tangent line.
For the function \( f(x) = \frac{8}{x} \), calculate the values at \( x = 5 \) and \( x = 8 \):
- \( f(5) = \frac{8}{5} \)
- \( f(8) = 1 \)
The slope of the secant line through the points \((5, f(5))\) and \((8, f(8))\) is given by:
\[
\text{slope} = \frac{f(8) - f(5)}{8 - 5} = \frac{1 - \frac{8}{5}}{3} = -\frac{1}{5}
\]
Using the point-slope form of a line, the equation of the secant line is:
\[
y - \frac{8}{5} = -\frac{1}{5}(x - 5)
\]
Simplifying, we get:
\[
y = -\frac{1}{5}x + \frac{13}{5}
\]
The derivative of \( f(x) = \frac{8}{x} \) is:
\[
f'(x) = -\frac{8}{x^2}
\]
Evaluate the derivative at \( x = 5 \):
\[
f'(5) = -\frac{8}{25}
\]
Using the point-slope form of a line, the equation of the tangent line at \( x = 5 \) is:
\[
y - \frac{8}{5} = -\frac{8}{25}(x - 5)
\]
Simplifying, we get:
\[
y = -\frac{8}{25}x + \frac{16}{5}
\]
- The equation of the secant line is: \(\boxed{y = -\frac{1}{5}x + \frac{13}{5}}\)
- The equation of the tangent line is: \(\boxed{y = -\frac{8}{25}x + \frac{16}{5}}\)
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