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f(X) = 2x^2 + 4 at x=1 Find the derivative, both a and b should have the same answer.

 

a.) Use equation lim as x -> c f(x)-f(c)/x-c 

 

b.) Use equation lim as h ->0 f(c+h)-f(c)/h

Yura_chan  Feb 4, 2016

Best Answer 

 #1
avatar+90988 
+15

Hi Yura_chan,

 

I've done the algebra below but you reall want to graph this and try to understand what is happening :/

 

f(X) = 2x^2 + 4 at x=1 Find the derivative, both a and b should have the same answer.

 

a.) Use equation lim as x -> c f(x)-f(c)/x-c 

 

\(\displaystyle \lim_{x\rightarrow1} \;\;\frac{f(x)-f(1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2x^2+4-6}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2x^2-2}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x^2-1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x-1)(x+1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x+1)}{1}\\ =2*(1+1)\\ =4\)

 

b.) Use equation lim as h ->0 f(c+h)-f(c)/h

 

\(\displaystyle \lim_{h\rightarrow0} \;\;\frac{f(c+h)-f(c)}{c+h-c}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{f(1+h)-f(1)}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h^2+4h+6-6}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h^2+4h}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h(h+2)}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2(h+2)}{1}\\ =2*(0+2)\\ =4\)

 

 

this means that the gradient of the tangent to the curve f(x)=2x^2+4        is     4   when x=1

Melody  Feb 4, 2016
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3+0 Answers

 #1
avatar+90988 
+15
Best Answer

Hi Yura_chan,

 

I've done the algebra below but you reall want to graph this and try to understand what is happening :/

 

f(X) = 2x^2 + 4 at x=1 Find the derivative, both a and b should have the same answer.

 

a.) Use equation lim as x -> c f(x)-f(c)/x-c 

 

\(\displaystyle \lim_{x\rightarrow1} \;\;\frac{f(x)-f(1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2x^2+4-6}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2x^2-2}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x^2-1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x-1)(x+1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x+1)}{1}\\ =2*(1+1)\\ =4\)

 

b.) Use equation lim as h ->0 f(c+h)-f(c)/h

 

\(\displaystyle \lim_{h\rightarrow0} \;\;\frac{f(c+h)-f(c)}{c+h-c}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{f(1+h)-f(1)}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h^2+4h+6-6}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h^2+4h}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h(h+2)}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2(h+2)}{1}\\ =2*(0+2)\\ =4\)

 

 

this means that the gradient of the tangent to the curve f(x)=2x^2+4        is     4   when x=1

Melody  Feb 4, 2016
 #2
avatar+209 
+5

Thank you! I'm not allowed to use a calculator to graph on the test and I don't really understand the graph aspect of this yet...but this is what I needed to show on the problem, so thanks!

I will definitely be looking into the graphing though, as that is something I really struggle with.

Yura_chan  Feb 4, 2016
 #3
avatar+90988 
+10

I understand what you are saying Yura but I would love to be teaching you face to face right now.

 

I know you are expected to memorize the formula and I understand that formula are nice and comfortable and if you memorize them they work really well.

I too learned formula for this before I really understood what was happening.

 

BUT if you understood what was happening you would not need the formula - the method would just be obvious.

 

I know there is very little time for anything when you studying conscienciously full time but try to make time to understand graphs.  If you understand them then many other things become so much easier.  Many things just become obvious!  laugh

 

Here is an introduction to derivatives that I'd like you to take a look at.

It is not exactly what I was looking for but it may serve you well anyway.  :/

http://www.mathsisfun.com/calculus/derivatives-introduction.html

 

 

THE MOST IMPORTANT THING TO REMEMBER IS THAT THE DERIVATIVE IS THE GRADIENT OF THE TANGENT TO A CURVE AT A GIVEN POINT!    laugh laugh laugh

Melody  Feb 6, 2016
edited by Melody  Feb 6, 2016

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