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I'm not sure how to go on and answer this question, I appreciate any help.

Show that $$f(x) = \ln(3x^2 - 2x -1) - 4x^2$$ has a stationary value when $x$ satisfies $$12x^3 - 8x^2 - 7x + 1 =0$$

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2 Answers 2

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To find a stationary value, one must first find the first derivative of $f(x)$. In doing so you should find $$f'(x) = \frac{6x-2}{3x^{2}-2x-1}-8x$$ For the left hand side to be zero (i.e., a stationary value) $$0=\frac{6x-2}{3x^{2}-2x-1}-8x$$ Implying $$6x-2=8x(3x^{2}-2x-1)$$ Multiply this out and re-arrange and you should find what you are looking for.

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$$f'(x)=\frac{6x-2}{3x^2-2x-1}-8x=0$$

rearrange to get the given equation

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