Question

A tangent line and its normal line have a point of tangency to the function f(x) at (x,y). If the slop of the normal line is m=(8/7), what are the coordinates of the point of tangency?
Please help, step by step explanation would be great

Answer 1

Answer:

$\displaystyle \\\left(\frac{49}{900},\frac{3761}{900}\right)$ or approximately (0.544, 4.179)

Step-by-step explanation:

A function and its tangent lines intersect when their slopes are the same. Find the x-coordinate when the slope of f(x) is equal to 8/7 by taking the derivative of f(x):

$\displaystyle\\f(x)=\sqrt{x}-x+4\\f'(x)=\frac{1}{2}\cdot\frac{1}{\sqrt{x}}-1=\frac{1}{2\sqrt{x}}-1$

Set $f'(x)$ equal to 8/7 and solve for x:

$\displaystyle \\\frac{1}{2\sqrt{x}}-1=\frac{8}{7},\\x=\frac{49}{900}$

Therefore, $f(x)$ will intersect at a point of tangency with a line of slope 8/7 at x=49/900. Plug in x=49/900 into $f(x)$ to get the y-coordinate:

$\displaystyle\\y=\sqrt{x}-x+4 \vert_{x=49/900}=\frac{3761}{900}$

⇒Answer: (49/900, 3761/900) or approximately (0.544, 4.179)