Question

I have a calculus question about limit functions. Pic included.

Answer 1

Answer:

Based on the results above, the slope of the tangent line to the curve at P(25,10) is 0.1.

Explanation:

P(25, 10) lies on the curve:

$y=\sqrt{x}+5$

Q is the point (x, √x+5).

For any value of x, the slope of the secant line PQ is determined using the formula:

$\text{ Slope of secant PQ}=\frac{(\sqrt{x}+5)-10}{x-25}$

(a)When x=25.1

$\text{Slope of secant PQ}=\frac{(\sqrt{25.1}+5)-10}{25.1-25}=0.0999$

(b)When x=25.01

$\text{Slope of secant PQ}=\frac{(\sqrt{25.01}+5)-10}{25.01-25}=0.09999$

(c)When x=24.9

$\text{Slope of secant PQ}=\frac{(\sqrt{24.9}+5)-10}{24.9-25}=0.1001$

(d)When x=24.99

$\text{Slope of secant PQ}=\frac{(\sqrt{24.99}+5)-10}{24.99-25}=0.10001$

Based on the results above, the slope of the tangent line to the curve at P(25,10) is 0.1.