randy17 Posted June 3, 2013 Share Posted June 3, 2013 The problem statement, all variables and given/known dataFind the critical point of f(x)=(-5x^2+3x)/(2x^2-5), and show whether this point is a max/min. Relevant equationsf(x)=(-5x^2+3x)/(2x^2-5)The attempt at a solutionWhen I tried solving for the derivative of; f(x)=(-5x^2+3x)/(2x^2-5), I got (-6x^2+50x-15)/2x^2-5)^2. then I set it equal to 0 and I got 0=-6x^2+50x-15Now from here do I use the quadratic formula to solve for x? I tried that and I am getting x=(25+sqrt535)/6 and x=(25-sqrt535)/6. Have I done it right so far? When I try to plug in the x-values into the original eq to get y my calc says error... What do I do now? Link to comment Share on other sites More sharing options...
mathmari Posted June 3, 2013 Share Posted June 3, 2013 (edited) Yes, you has done it right... For [latex] x<=(25-\sqrt{535})/6 [/latex], [latex]f'(x)<=0 [/latex]For [latex] (25-\sqrt{535})/6<x<(25+\sqrt{535})/6 [/latex], [latex]f'(x)>0 [/latex]For [latex] x>=(25+\sqrt{535})/6 [/latex], [latex] f'(x)<=0 [/latex] (The domain of the function is [latex]R[/latex] \ { [latex]-\sqrt{\frac{5}{2}}, \sqrt{\frac{5}{2}} [/latex] } ) So the min of this function is at [latex] x=(25-\sqrt{535})/6 [/latex], and it's[latex] f((25-\sqrt{535})/6)=(\sqrt{535}-25)/20[/latex]And the max is at [latex] x=(25+\sqrt{535})/6[/latex], [latex] f((25+\sqrt{535})/6)=-(\sqrt{535}+25)/20[/latex] Edited June 3, 2013 by mathmari Link to comment Share on other sites More sharing options...
daniton Posted June 3, 2013 Share Posted June 3, 2013 Why don't you just second derivative test.It's easy and quick. Link to comment Share on other sites More sharing options...
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