Respuesta :
The question is incomplete. The complete question is:
Find an equation of the tangent line to the curve y = [tex]\sqrt{x}[/tex] at the given point (16,4). To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equation of the tangent line at the point (16,4) we know that (16,4) is a point on the line. So we just need to find its slope. The slope of a tangent line to f(x) at x = a can be found using the formula m tan = lim x↔a f(x) - f(a)/ x - a.
Answer: y = [tex]\frac{x}{8} + 2[/tex]
Step-by-step explanation: The tangent line is a line that intercepts a curve in only one point. The point-slope formula for a line is [tex]y-y_{0} = m(x-x_{0})[/tex], where m is the slope of the line and can be calculated by the first derivative of the given curve. For this case:
y = [tex]\sqrt{x}[/tex]
f'(x) = [tex]\frac{dy}{dx} = \sqrt{x}[/tex]
f'(x) = [tex]\frac{1}{2\sqrt{x} }[/tex]
At point (16,4), the slope will be:
m = f'(16) = [tex]\frac{1}{2.\sqrt{16} }[/tex]
m = [tex]\frac{1}{8}[/tex]
With slope and a point, the line function will be:
[tex]y-y_{0} = m(x-x_{0})[/tex]
y - 4 = [tex]\frac{1}{8}[/tex](x - 16)
y = [tex]\frac{x}{8}[/tex] - 2 + 4
y = [tex]\frac{x}{8}[/tex] + 2
The tangent line to the curve is y = x/8 + 2
