Let f be the function given by f(x)=3x^4 +x^3 -21x^2
a.) Write an equation of the line tangent to the graph of f at the point (2,-28) b.) Find the absolute minimum value of f. Show the analysis that leads to your conclusion c.) Find the x-coordinate of each point of inflection on the graph of f. Show the analysis that leads to your conclusions
@phi
this looks like calculus for a) find the derivative of your curve, and sub in x=2 to find the slope at that point. once you have the slope, use the point slope formula for the equation of the tangent line.
kay i got y=24x-76
How about for part b.) and c.)
@phi
ok on a) for the min, set the derivative = 0 and solve for x see http://www.khanacademy.org/math/calculus/derivative_applications/critical_points_graphing/v/minima--maxima-and-critical-points it is short and helpful
@phi so my answer for a.) y=24x-76 is wrong?
or did u mean b.) not a.)
I said you were ok on a). That means your answer for a is correct. the video is to help you finish the problem.
would it be easier if i divided the derivative by 3: 12x^3 +3x^2 - 42x=0?
of course, also factor out an x (x=0 is one solution)
so ishould factor out an x and 3 o: 3x(4x^2 +x - 14)?
3x(4x^2 +x - 14)=0
ok so i got 3x=0 (4x-7)(1x+2)=0 x=0, 7/4, 2
@phi
you mean x=-2
time to watch the video, and maybe the one after it where he does this same type of problem
oh yeah sorry i mean -2 :)
sorry i can't really watch the vid cause my cp is lagging x-x
so would x=-2 be the absolute minimum value of f?
@phi
ok how about for part c.)
@agent0smith can you help me with part c.) please? :)
Part c just wants you to find every point where the gradient is zero. So, wherever f'(x) is equal to zero.
Which...it looks like you already did. x=0, 7/4, -2
ok thank you can u help mewith another please ^^
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