Critical Number Problem - # 5

[Jason76]

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8\)

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x^{2} + 3x - 9) = 0\)

\(\displaystyle -(3) \pm \sqrt{\dfrac{3^{2} - 4(2)(-9)}{(2)(2)}}\)

\(\displaystyle -(3) \pm \sqrt{\dfrac{81}{4}}\)

\(\displaystyle -(3) \pm \sqrt{15.75}\) Something is wrong somewhere in this problem.

 

[stapel]

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8\)

What were the instructions for this function? What are you supposed to be doing with it?

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x^{2} + 3x - 9) = 0\)

\(\displaystyle -(3) \pm \sqrt{\dfrac{3^{2} - 4(2)(-9)}{(2)(2)}}\)

Had you taken algebra before attempting calculus, you would have encountered the Quadratic Formula. Try applying that to the equation, rather than whatever it is that you're using in the last line above.

\(\displaystyle -(3) \pm \sqrt{15.75}\) Something is wrong somewhere in this problem.

Among other issues, there is no justification for using decimals rather than the exact form. Also, you might want to consider learning your multiplication table, so you can recognize square numbers. For instance, the square root of four is simply two.

 

[Deleted member 4993]

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8\)

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x^{2} + 3x - 9) = 0\)

\(\displaystyle -(3) \pm \sqrt{\dfrac{3^{2} - 4(2)(-9)}{(2)(2)}}\) ← That should be:

\(\displaystyle \displaystyle x_{1,2} \ = \ \dfrac{-3\ \pm \ \sqrt{(3)^2 - 4*(2)*(-9)}}{2*2}\)

\(\displaystyle -(3) \pm \sqrt{\dfrac{81}{4}}\)

\(\displaystyle -(3) \pm \sqrt{15.75}\) Something is wrong somewhere in this problem.

.

 

[Jason76]

.

Ok, got it, minor error. Square root only on the top, no encompassing everything to the right of the equals sign.

 

[srmichael]

Ok, got it, minor error. Square root only on the top, no encompassing everything to the right of the equals sign.

Minor error??? Coloring outside the lines with a crayon is a minor error. This is just flat out incorrect. Learn the formula, apply it correctly and then all will be right as rain.

 

[Jason76]

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8\)

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x^{2} + 3x - 9) = 0\)

\(\displaystyle -(3) \pm \dfrac{\sqrt{3^{2} - 4(2)(-9)}}{(2)(2)}\)

\(\displaystyle -(3) \pm \dfrac{\sqrt{45}}{(2)(2)}\)

\(\displaystyle -(3) \pm \dfrac{\sqrt{9} * \sqrt{5}}{(2)(2)}\)??

 

[JeffM]

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8\)

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x^{2} + 3x - 9) = 0\)

\(\displaystyle -(3) \pm \dfrac{\sqrt{3^{2} - 4(2)(-9)}}{(2)(2)}\)

\(\displaystyle -(3) \pm \dfrac{\sqrt{45}}{(2)(2)}\)

\(\displaystyle -(3) \pm \dfrac{\sqrt{9} * \sqrt{5}}{(2)(2)}\)??

Jason

You have been told repeatedly that you are getting the Quadratic Formula wrong.

Plus how do you figure that 9 + 72 = 45.

Review SK's post above.

Edit: By the way, if you derive the formula yourself, you are far less likely to get it wrong in the future

 

[lookagain]

Determine the critical numbers of \(\displaystyle \ f(x) = 4x^{3} + 9x^{2} - 54x + 8. \ \ \ \ \ \ \) A

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x^{2} + 3x - 9) = 0\)


\(\displaystyle \dfrac{6(2x^{2} + 3x - 9)}{6} = \dfrac{0}{6}\)


\(\displaystyle 2x^2 + 3x - 9 = 0 \ \ \ \ \ \ \)B


\(\displaystyle x \ = \ \dfrac{-b \pm \sqrt{b^2 - 4ac \ }}{2a} \ \ \ \ \ \ \) C


\(\displaystyle x \ = \ \dfrac{-(3) \pm \sqrt{(3)^{2} - 4(2)(-9) \ }}{2(2)} \ \ \ \ \ \ \)D


\(\displaystyle x \ = \ \dfrac{-3 \pm \sqrt{ \ 9 + 72 \ }}{4} \ \ \ \ \ \ \)E

\(\displaystyle x \ = \ \dfrac{-3 \pm \sqrt{ \ 81 \ }}{4}\)


Continue . . .


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Jason76,

A) State the instructions in the body of your post, not just some supposed non-clarified instructions in the headline/heading.


B) Write the the newest equation on which you're using the Quadratic Formula.


C) Write out the Quadratic Formula as a model. In this context, *this includes* wriiting "x =" in front of the "large fraction"
with the radical, because you are just transforming one equation into another equation.

D) Use parentheses (or other grouping symbols) to show the values you are substituting for in the Quadratic Formula.

E) Be careful of your arithmetic when evaluating the radicand.

 

[Jason76]

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8\)

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x^{2} + 3x - 9) = 0\)

\(\displaystyle \dfrac{-(3) \pm \sqrt{3^{2} - 4(2)(-9)}}{2(2)}\)

\(\displaystyle \dfrac{-(3) \pm \sqrt{81}}{4}\)

\(\displaystyle \dfrac{-(3) \pm 9}{4}\)

\(\displaystyle \dfrac{-(3) + 9}{4} = \dfrac{6}{4} = \dfrac{3}{2} \)

\(\displaystyle \dfrac{-(3) - 9}{4} = -\dfrac{12}{4} = -3 \)

 

[lookagain]

\(\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8\)

\(\displaystyle f'(x) = 12x^{2} + 18x - 54\)

\(\displaystyle 12x^{2} + 18x - 54 = 0\)

\(\displaystyle 6(2x^{2} + 3x - 9) = 0\)

\(\displaystyle \dfrac{-(3) \pm \sqrt{3^{2} - 4(2)(-9)}}{2(2)} \ \ \ \)<------This and the next four steps are wrong, at least, for missing "x =."

\(\displaystyle \dfrac{-(3) \pm \sqrt{81}}{4} \ \ \ \)<------

\(\displaystyle \dfrac{-(3) \pm 9}{4} \ \ \ \)<------

\(\displaystyle \dfrac{-(3) + 9}{4} = \dfrac{6}{4} = \dfrac{3}{2} \ \ \ \) <-------

\(\displaystyle \dfrac{-(3) - 9}{4} = -\dfrac{12}{4} = -3 \ \ \ \) <------

Jason76, why ignore certain corrections and *bother* typing wrong stuff!? You don't

have correct answers. Part of the correct work, as I *already stated*, is that solutions to the

quadratic equation(s) (with x as the variable), such as using the Quadratic Formula, contain

"x = " as part of the answer.