Answer:
[tex]m = \frac{1}{12}[/tex]
Step-by-step explanation:
Given
[tex](x,y) = (36,6)[/tex]
[tex]f(x) = \sqrt x[/tex] ----- the equation of the curve
Required
The slope of f(x)
The slope (m) is calculated using:
[tex]m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}[/tex]
[tex](x,y) = (36,6)[/tex] implies that:
[tex]a = 36; f(a) = 6[/tex]
So, we have:
[tex]m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}[/tex]
[tex]m = \lim_{h \to 0} \frac{f(36 + h) - 6}{h}[/tex]
If [tex]f(x) = \sqrt x[/tex]; then:
[tex]f(36 + h) = \sqrt{36 + h}[/tex]
So, we have:
[tex]m = \lim_{h \to 0} \frac{\sqrt{36 + h} - 6}{h}[/tex]
Multiply by: [tex]\sqrt{36 + h} + 6[/tex]
[tex]m = \lim_{h \to 0} \frac{(\sqrt{36 + h} - 6)(\sqrt{36 + h} + 6)}{h(\sqrt{36 + h} + 6)}[/tex]
Expand the numerator
[tex]m = \lim_{h \to 0} \frac{36 + h - 36}{h(\sqrt{36 + h} + 6)}[/tex]
Collect like terms
[tex]m = \lim_{h \to 0} \frac{36 - 36+ h }{h(\sqrt{36 + h} + 6)}[/tex]
[tex]m = \lim_{h \to 0} \frac{h }{h(\sqrt{36 + h} + 6)}[/tex]
Cancel out h
[tex]m = \lim_{h \to 0} \frac{1}{\sqrt{36 + h} + 6}[/tex]
[tex]h \to 0[/tex] implies that we substitute 0 for h;
So, we have:
[tex]m = \frac{1}{\sqrt{36 + 0} + 6}[/tex]
[tex]m = \frac{1}{\sqrt{36} + 6}[/tex]
[tex]m = \frac{1}{6 + 6}[/tex]
[tex]m = \frac{1}{12}[/tex]
Hence, the slope is 1/12
