Find k such that the line is tangent to the graph of the function.

Question

Function: f(x)= kx

Line: y=4x+4

Answer 1

There are two conditions the the line to be tangent to the function. 1) the function and the line must intersect. 2) the slope of the tangent line of the function must be the same as the slope of the line y = 4x + 4

For condition 1: If y = k√x and y = 4x + 4 then k√x = 4x + 4 or k = (4x + 4)/√x

For condition 2: if f(x) = k√x, we can write it as f(x) = k(x)^1/2 and using the power rule, we can take the derivative so f ' (x) = 1/2k(x)^-1/2 or k/(2√x) and that derivative must equal the slope of the line y = 4x + 4 (or 4) which means k/(2√x) = 4 or k = 4(2√x) or k = 8√x

So, putting conditions 1 and 2 together, k = (4x + 4)/√x and k = 8√x we can say that:

(4x + 4)/√x = 8√x

(4x + 4) = 8√x(√x)

4x + 4 = 8x

4 = 4x

x = 1

since (from condition 2) k = 8√x then k = 8√1 or k = 8

Answer 2

First of all, realize that k>0 since the slope of the given line is positive. Otherwise, the graph square root would be reflected with respect to the x-axis and will never be tangent to the line.

Next, the slope at every point $x>0$ (because the domain of the square root x^{1/2} is all positive real numbers) is given by f'(x) = k/(2sqrt(x)). We want this slope to be as the one of the given line. That is, we want

k/(2sqrt(x)) = 4

which implies that k/8 = sqrt(x) and thus x = k^2/64. Next, notice that f(k^2/64) = k*sqrt(k^2/64) = k^2/8. Thus, the line must pass through the point (k^2/64, k^2/8). In other words,

4 (k^2/64) + 4 = k^2/8

And we solve this equation to obtain k=8. Recall x>0!