Respuesta :
slope = [tex]\frac{1}{12}[/tex]
the slope is the value of f' (36)
f(x) = √x = [tex]x^{\frac{1}{2} }[/tex]
f'(x) = [tex]\frac{1}{2}[/tex] [tex]x^{-\frac{1}{2} }[/tex] = [tex]\frac{1}{2\sqrt{x} }[/tex]
f'(36) = [tex]\frac{1}{2(6)}[/tex] = [tex]\frac{1}{12}[/tex]
Answer:
[tex]slope = \frac{1}{12}[/tex]
Step-by-step explanation:
Here is another method to solve your problem. I am showing this method because this is the first method normally taught and a student might not of had the chance yet to learn the other methods
We can solve this problem by using limits and the following function
[tex]\lim_{h\to 0} \frac{f(x+h) - x}{h}[/tex]
[tex]\lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}[/tex]
Next multiply by the conjugate of the numerator.
[tex]\lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} * \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}[/tex]
[tex]\lim_{h\to 0} \frac{x + h - x}{h(\sqrt{x+h} + \sqrt{x})}[/tex]
Cancel the x - x
[tex]\lim_{h\to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}[/tex]
Divide out the h
[tex]\lim_{h\to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}[/tex]
[tex]\lim_{h\to 0} \frac{1}{(\sqrt{x+h} + \sqrt{x})}[/tex]
Plugin 0 where h is located
[tex]\lim_{h\to 0} \frac{1}{(\sqrt{x+h} + \sqrt{x})}[/tex]
[tex]\lim_{h\to 0} \frac{1}{(\sqrt{x+0} + \sqrt{x})}[/tex]
[tex]\lim_{h\to 0} \frac{1}{(\sqrt{x} + \sqrt{x})}[/tex]
Combine Like terms in denominator
[tex]\lim_{h\to 0} \frac{1}{(\sqrt{x} + \sqrt{x})}[/tex]
[tex]\lim_{h\to 0} \frac{1}{2\sqrt{x}}[/tex]
Now lets use our derivative and plugin 36 where x is located and solve
[tex]\frac{1}{2\sqrt{x}}[/tex]
[tex]\frac{1}{2\sqrt{36}}[/tex]
[tex]\frac{1}{2(6)}[/tex]
[tex]\frac{1}{12}[/tex]
Note, this is a harder method but it is normally the first method taught in Calculus 1.
