Aliyah Thompson

2022-11-23

For f(x)=x^4-12x^3+6 find the following.
A)f'(x)
B)The slope of the graph of f at x=-2
C)The equation of the tangent line at x=-2
D)The value(s) of x where the tangent line is horizontal

Aliya Moore

Expert

Given $f\left(x\right)={x}^{4}-12{x}^{3}+6$
A) ${f}^{\prime }\left(x\right)=4{x}^{3}-36{x}^{2}+0\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)=4{x}^{3}-36{x}^{2}$
Answer: ${f}^{\prime }\left(x\right)=4{x}^{3}-36{x}^{2}$
B)Slope of graph of f at x=-2
${f}^{\prime }\left(-2\right)=4\left(-2{\right)}^{3}-36\left(-2{\right)}^{2}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(-2\right)=-32-144\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(-2\right)=-176$
C) at x=-2
$y=f\left(-2\right)=\left(-2{\right)}^{4}-12\left(-2{\right)}^{3}+6\phantom{\rule{0ex}{0ex}}y=f\left(-2\right)=16+96+6\phantom{\rule{0ex}{0ex}}y=118$
equation of tangent line
$y-{y}_{1}=m\left(x-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}y-118=-176\left(x+2\right)\phantom{\rule{0ex}{0ex}}y-118=-176x-352\phantom{\rule{0ex}{0ex}}y=-176x-234$
D)Since tengent line is horizontal,
Hence

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