# Re: applied calculus

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Posted by Marilyn on September 28, 2002 at 17:00:47:

In Reply to: Re: applied calculus posted by Jack on September 28, 2002 at 10:53:23:

: : I am really struggling. This is not my expertise. Can someone help me understand this? I have a test on Tuesday and need all the input I can get.
: : #1) Find the equation of the tangent line to the curve f(x)= x^3 - x, at x= -1
: : The derivative of a function at a point is the slope of the tangent line at this point. So find f'(x)=3x^2-1. f'(x=-1)=3(-1)^2-1=2 which is the slope of the tangent line. Also, f(x=-1)=(-1)^3-(-1)=0. Recall that the standard form of the equation for a line is y=mx+b. We can find the value of b by subbing in what we know. m=2, x=-1, y=0, so b=-mx=-(2)(-1)=2. Thus, the equation of the tangent line through x=-1 is y=2x+2.

: : #2) Find the equation of the tangent line to the curve f(x) = x3 - x2 + 1 at the point (1,1).
: : As in #1, find f'(x)=3x2-2x or slope=m=f'(x=1)=3-2=1. Using y=mx+b, sub in what is known to find b. b=y-mx=1-(1)(1)=0. The equation of the tangent line is y=x.

: : #3) If f(x) = 2-3x^2 / x^3 + x - 1, find f'(x).
: : First, simplify f(x) and get f(x)=1-3x^(-1)+x
: :f'(x)=0-3(-1)x^(-2)+1= 1+3/x^2

: : #4) Find the derivative when f(x)= 5x^2 - 7x + 1. (Use the difference quotient.)
: :I am not aware of a "difference quotient" rule. I will use the "sum rule" and "difference rule". Lets write f(x)=d(x)-g(x)+h(x) where d(x)=5x^2, g(x)=7x and h(x)=1. Thus, f'(x)=d'(x)-g'(x)+h'(x). Now, d'(x)=(5)(2)x^(2-1)=10x. g'(x)=(7)(1)x^(1-1)=7. h'(x)=0. So we have f'(x)=10x-7+0=10x-7.

: : If someone could show step by step in solving, maybe I will get it? Need response as soon as possible.

: If you need further help, please e-mail me.

: : Thanks

Thanks for your help!

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