To find the value of k for which the line y = 6x + k is tangent to the curve y = 7/√x, we need to equate the derivatives and solve for k.

The derivative of y = 7/√x is:

\(\frac{d}{dx} \left( 7x^{-1/2} \right) = -\frac{7}{2}x^{-3/2}\)

The derivative of y = 6x + k is:

\(\frac{d}{dx} (6x + k) = 6\)

Setting the derivatives equal gives:

\(-\frac{7}{2}x^{-3/2} = 6\)

Solving for x:

\(x = \frac{49}{144}\)

Substitute x back into the curve equation to find y:

\(y = \frac{49}{12}\)

Substitute x and y into the line equation:

\(\frac{49}{12} = 6 \left( \frac{49}{144} \right) + k\)

Solve for k:

\(k = \frac{49}{24} \text{ or } 2.04\)