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son4ous [18]
2 years ago
12

4. The slope of the tangent for the function y = 1x is 1/(2(x). Find the equation of the tangent line at the point x = 1. Illust

rate on a graph.

Mathematics
1 answer:
Marianna [84]2 years ago
3 0

Answer:

x-2y+1=0    

Step-by-step explanation:

We are given the following information in the question:

y = \sqrt{x}

Differentiating y with respect to x:

\displaystyle\frac{dy}{dx} = \frac{1}{2\sqrtx}

At x = 1

\displaystyle\frac{dy}{dx}\Bigr|_{\substack{x=1} }= \frac{1}{2\sqrt{1}}=\frac{1}{2}

y(1) = \sqrt{1} = 1

Equation of tangent:

(y-y_0) = \displaystyle\frac{dy}{dx}(x-x_0)

Putting the values:

(y-y(1)) = \displaystyle\frac{dy}{dx}\Bigr|_{\substack{x=1} }(x-1)\\\\y - 1= \frac{1}{2}(x-1)\\2y -2 = x - 1\\x-2y+1=0  

The above equations are the required equation of the tangent.

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