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# Quick little explanation of easy equation of tangent like

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415
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+280

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.

Oct 19, 2017

#1
+7350
+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 + $$\frac52$$

.
Oct 19, 2017

#1
+7350
+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
#2
+98125
+1

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

Oct 19, 2017
#3
+7350
+1

Thank you! And thanks for your help!!

hectictar  Oct 19, 2017