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 I need to be able to know how to do (3.) for an exam next week. I just dont know if i put that function into the one above or do i just solve it. so i would get 0 for the slope.

Veteran  Oct 19, 2017

Best Answer 

 #1
avatar+5256 
+1

g(x)  =  \(5-\sqrt{4-x}\)

 

When  x = 3 , the slope of the curve  \(=\,\lim\limits_{h\to0}\frac{g(3+h)-g(3)}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{(5-\sqrt{4-(3+h)})-(5-\sqrt{4-3})}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{5-\sqrt{4-3-h}-4}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{1-\sqrt{1-h}}{h} \\~\\ =\,\lim\limits_{h\to0}(\,\frac{1-\sqrt{1-h}}{h}\,)\,(\,\frac{1+\sqrt{1-h}}{1+\sqrt{1-h}}\,) \\~\\ =\,\lim\limits_{h\to0}\frac{1-(1-h)}{h+h\sqrt{1-h}} \\~\\ =\,\lim\limits_{h\to0}\frac{h}{h+h\sqrt{1-h}} \\~\\ =\,\lim\limits_{h\to0}\frac{1}{1+\sqrt{1-h}} \\~\\ =\,\frac{1}{1+\sqrt{1-0}} \\~\\ =\,\frac12\)

 

 

Look at the graph and see that it does look like that line has a slope of \(\frac12\) .

 

We want an equation of a line that has a slope of  \(\frac12\)  and passes through the point  ( 3, g(3) ) .

 

g(3)  =  \(5-\sqrt{4-3}\)  =  5 - 1  =  4

 

So the equation of our line in point - slope form is

 

y - 4  =  \(\frac12\)(x - 3)

 

And in slope - intercept form, this is...     y  =  \(\frac12\)x + \(\frac52\)

hectictar  Oct 19, 2017
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3+0 Answers

 #1
avatar+5256 
+1
Best Answer

g(x)  =  \(5-\sqrt{4-x}\)

 

When  x = 3 , the slope of the curve  \(=\,\lim\limits_{h\to0}\frac{g(3+h)-g(3)}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{(5-\sqrt{4-(3+h)})-(5-\sqrt{4-3})}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{5-\sqrt{4-3-h}-4}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{1-\sqrt{1-h}}{h} \\~\\ =\,\lim\limits_{h\to0}(\,\frac{1-\sqrt{1-h}}{h}\,)\,(\,\frac{1+\sqrt{1-h}}{1+\sqrt{1-h}}\,) \\~\\ =\,\lim\limits_{h\to0}\frac{1-(1-h)}{h+h\sqrt{1-h}} \\~\\ =\,\lim\limits_{h\to0}\frac{h}{h+h\sqrt{1-h}} \\~\\ =\,\lim\limits_{h\to0}\frac{1}{1+\sqrt{1-h}} \\~\\ =\,\frac{1}{1+\sqrt{1-0}} \\~\\ =\,\frac12\)

 

 

Look at the graph and see that it does look like that line has a slope of \(\frac12\) .

 

We want an equation of a line that has a slope of  \(\frac12\)  and passes through the point  ( 3, g(3) ) .

 

g(3)  =  \(5-\sqrt{4-3}\)  =  5 - 1  =  4

 

So the equation of our line in point - slope form is

 

y - 4  =  \(\frac12\)(x - 3)

 

And in slope - intercept form, this is...     y  =  \(\frac12\)x + \(\frac52\)

hectictar  Oct 19, 2017
 #2
avatar+78755 
+1

Very nice, hectictar........!!!!

 

 

cool cool cool  

CPhill  Oct 19, 2017
 #3
avatar+5256 
+1

Thank you! And thanks for your help!! laugh

hectictar  Oct 19, 2017

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