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Find k such that the line is tangent to the graph of the function?
Function: f(x) = kx3
Line: y = 8x + 7
I keep getting -21/16... don't know what i'm doing wrong
1 Answer
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- DWReadLv 74 years ago
-21/16 is not the value of k. It is the value of x at the point of tangency.
f(-21/16) = k(-21/16)³
k(-21/16)³ = 8(-21/16) + 7
solve for k
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