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Question: Find the maximum y value of the graph between 0 ≤ x ≤ 3
Equation:
f(x) = x(8-2x)(6-2x)
Answer:
24.26
How would you find the maximum y value of this graph? Is it not like a parabola so when I tried finding the midpoint and then using the quadratic formula it didn't give me the highest value. Please explain the process and show your work. Thanks.
Equation:
f(x) = x(8-2x)(6x-2x)??
Oops typo.
Corrected:
f(x) = x(8-2x)(6-2x)
Equation:
f(x) = x(8-2x)(6-2x) = 4x(4-x)(3-x) = $\displaystyle 4x^{3}-28x^{2}+48x$
at the turning points
$\displaystyle d/dx = 0$
$\displaystyle d/dx = 12x^{2}-56x+48$
roots are $\displaystyle (7-sqr(13))/3$ and $\displaystyle (7+sqr(13))/3$
until now all we know is these are turning points,
The value f ”(x) will tell us whether the point is a maximum or a minimum or a point of inflection
differentiate again
f''(x)=24x-56= 8(3x-7)
f''((7+sqr(13))/3) = positive, hence it is min
f''((7-sqr(13))/3) = negative, so this is max
f((7-sqr(13))/3) = 24.26
Answer:
24.26
you also need to check the global extrema, where the graph gets cut off e.g. at where x=3 and x=0 however in this case they are less than this value
