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conCentrate9
United States438 Posts
For the function f(x) = x^3 - 3x^2 + x + 4 find "k" such that f(x) has only one tangent line of slope "k" All I've done so far is taken the derivative to be f'(x) = 3x^2 - 6x + 1 I've just never seen a problem similar to this one and I'm stumped. Any ideas?   | ||
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Caller
Poland8075 Posts
k = 3x^2 -6x +1 solve for k for such k that there is only one solution. in other words find a discriminant that is 0 discriminant = b^2-4ac (-6)^2 - (4)(1)(k) find k such that the above expression = 0 = 36-4k k = 9 | ||
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Nitrogen
United States5345 Posts
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conCentrate9
United States438 Posts
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conCentrate9
United States438 Posts
b^2-4ac=0 36-4(3)(1-k)=0 36-12(1-k)=0 36-12+12k=0 24+12k=0 k = -2 Thanks. | ||
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Caller
Poland8075 Posts
ma bad | ||
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