Asked by sandy
Find the equation of the tangent line to y=f(x) at the point where at x=3.
f(x)=-4x^3-5x+7
f(x)=-4x^3-5x+7
Answers
Answered by
oobleck
The point is (3,-116)
f'(x) = -12x^2-5
f'(3) = -113
so the line is
y+116 = -113(x-3)
the graphs are at
www.wolframalpha.com/input/?i=plot+y%3D-4x%5E3-5x%2B7%2C+y%3D-113%28x-3%29-116+for+0%3C%3Dx%3C%3D4
f'(x) = -12x^2-5
f'(3) = -113
so the line is
y+116 = -113(x-3)
the graphs are at
www.wolframalpha.com/input/?i=plot+y%3D-4x%5E3-5x%2B7%2C+y%3D-113%28x-3%29-116+for+0%3C%3Dx%3C%3D4
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