The line 8x-y-8=0 is tangent to the graph of f(x)=(x+k)(x^2-1) at the point (1, 0). Find the value of k.

Answer 1

f^'(x) = x²−1 + 2x(x +k)

f^'(1) = 2(1+k)

The slope of the tangent line is m= f^'(1 )= 8

That is 2(1+k) = 8

k=3

Answer 2

When the line 8x-y-8=0 is tangent to our function f(x) at a point, that also means they occupy that same point and they share the same slope. Plugging (1,0) into our equation gives us 0=0, so that doesn't help. However, using the slopes/derivatives will prove useful. Our line turns into

y=8x-8

y' = 8

This slope has to equal our f '(x) at x=1

f(x) = (x+k)(x²-1) = -k - x + k x^2 + x^3

f '(x)= 0 -1 + 2*kx +3x²

f'(1)=8=-1+2*k+3

--> k=3

Answer 3

The problem seems to be poorly constructed or there is some explanation missing.

No matter what k is:

f(x)=(x+k)(x^2-1) will be undefined for x=1 (you can't divide by zero). So (1,0) isn't actually a part of the graph of f and so it doesn't make sense how you can talk about a line tangent to the graph there.

Could the denominator be x^2+1 ?

Or something else?