## Calculus 10th Edition

$k = 4$
To solve this problem, we need to identify point (x,y) where both the function and the equation of the tangent line intersect. To do this, we can equate both to produce the following: $$k\sqrt x = x + 4$$ By rewriting the equation, we can achieve the following: $$(k\sqrt x)^{2} = (x + 4)^{2}$$ $$k^{2}x = x^{2} + 8x + 16$$ $$0 = x^{2} + (8 - k^{2})x + 16$$ We can now solve for x using the quadratic formula: $$x = \frac{-b +/- (b^{2} - 4ac)^{\frac{1}{2}}}{2a}$$ However, an interesting corollary to the quadratic formula states that the expression inside the square root, b2−4ac, is positive when x has 2 solutions, 0 when it only has 1 solution, and negative when x has no solution in real numbers. Since we know that, by definition, the tangent line only intersects at one point (x,y) with the function f(x), we know that x can only have one solution and, as such: $$b^{2} - 4ac = 0$$ By plugging in the appropriate values, we get: $$(8-k^{2})^{2} - 4(1)(16) = 0$$ $$(8 - k^{2})^{2} = 64$$ $$((8 - k^{2})^{2})^{\frac{1}{2}} = (64)^{\frac{1}{2}}$$ $$8 - k^{2} = +/- 8$$ which gives us 2 possible solutions. However, we know that $+8$ would give us $k=0$ which is not an appropriate solution to the original function $f(x) = k\sqrt x$ (because it would mean that $f(x) = 0$), therefore we can only use $-8$: $$8 - k^{2} = -8$$ $$16 = k^{2}$$ $$+/-4 =k$$ Because we've arrived at 2 possible solutions, we need to check if both satisfy the original conditions. If $k = +/-4$, then $f(x) = +/-4\sqrt x$, $f'(x) = \frac{+/-2}{\sqrt x}$ and $$f'(x) = 1 = \frac{+/-2}{\sqrt x}$$ $$x = 4$$ Therefore, for functions $f_{1}(x) = 4\sqrt x$ and $f_{2}(x) = -4\sqrt x$, their respective solutions when $x = 4$ are (4, 8) and (4, -8) respectively. Using the second point to generate the equation of a tangent line when the slope is $1$: $$y +8 = 1(x -4)$$ does not result in the original equation of the tangent line. Therefore, only $k = 4$ satisfies the original conditions.