At what points on the graph of $f(x)=2x^3-9x^2-12x+5$ is the slope of the tangent line 12?

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Answer 1 · 2015-04-13T20:46:10

The derivative of a function gives us the slope of a tangent line for a specified value of $x$

$f(x) = 2x^3-9x^2-12x+5$
implies
$f'(x) = 6x^2-18x-12$

and we are asked to find when this is $=12$

$6x^2-18x-12 = 12$

$x^2-3x -4=0$

which factors as
$(x-4)(x+1)=0$

when $x=4$
$f(x) = 2(4)^3-9(4)^2-12(4)+5$
$=-59$

when $x=-1$
$f(-1)=2(-1)^3-9(-1)^2-12(-1)+5$
$=6$

So $f(x)$ has a tangent with slope $12$ at
$(4,-59)$
and
$(-1,6)$

At what points on the graph of f(x)=2x^3-9x^2-12x+5f(x)=2x^3-9x^2-12x+5 is the slope of the tangent line 12?