Find a cubic function [tex] f(x) = ax^3 - bx^2 + cx - d [/tex] that has a local maximum value of 40 at x = 1 and a local minimum valud of -68 at x = 4(adsbygoogle = window.adsbygoogle || []).push({});

Since x = 1 and x = 4 are the max and min, respectively, then f'(x) must equal 0 at x = 1 and x = 4. Therefore [tex] f'(x) = (x-1)(x-4) = x^2 - 5x +4 [/tex]

Using the original function [tex] f'(x) = 3ax^2 - 2bx +c = x^2 - 5x + 4[/tex]

Setting the terms equal to each other we get

[tex] 3ax^2 = x^2 \Rightarrow a = \frac{1}{3} [/tex]

[tex] -2bx = -5x \Rightarrow b = \frac{5}{2} [/tex]

[tex] c = 4 [/tex]

Substituting these values back into the original equation I get

[tex] f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 4 - d [/tex]

When I solve for d using x = 1 and f(x) = 40 I get

[tex] d = \frac{1}{3} - \frac{5}{2} - 36 = \frac{-229}{6} [/tex]

However, when I solve for d using x = 4 and f(x) = -68 I get

[tex] d = \frac{4^3}{3} - \frac{5(4^2)}{2} + 84 = \frac{226}{3} [/tex]

Any ideas where I went wrong, or if I was down the wrong track to start with?

Thanks

Jeff

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# Homework Help: Find a cubic function

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