Calculus questions for math

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12)Find an equation for the tangent line to the graph of f(x)=2x^2-2x+3 at the point where x=1

13)Let f(7)=0, f'(7)=14,g(7)=1 and g'(7)=1/7 find h(7) if h(x)=f(x)/g(x)

Nov 18th, 2014

11. f1(x)=2x+2sec^2x


take the derivative: 4x -2 (power rule) 
put 1 in for x, like given in problem. 
so... 4(1)-2 = 2 
that would be the slope of the line 
then to find the point: you have the x-value, 1 (given) 
you put 1 in for x in the origianl equation. 
2(1)^2 - 2(1) +3 = 2 - 2 +3 = 3 
so in point-slope form the answer is: 

y - 3 = 2(x - 1) 

y= 2x+1


h'(x)=[f'(x)*g(x)-g'(x)*f(x)]/[g(x)]^2 {quotient rule} 

So h'(7)=[f'(7)*g(7)-g'(7)*f(7)]/[g(7)]^2 {subbing in} 
Nov 18th, 2014

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