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Find an equation of the tangent line to the curve at the given point. y = sqrt(x), (1, 1).

Question:

Find an equation of the tangent line to the curve at the given point.

{eq}y = \sqrt{x}, \; (1, \; 1) {/eq}

Answer and Explanation:

The first thing to do is to calculate the slope of the desired tangent line by deriving {eq}f(x) {/eq} and then plugging in {eq}x=1 {/eq}:

{eq}\begin{align*} \displaystyle f(x)& = \sqrt{x}\\ f'(x)& =\frac{1}{2\sqrt{x}} \\ f'(1)& =\frac{1}{2\sqrt{1}} \\ f'(1)& =\frac{1}{2 } \\ \end{align*} {/eq}

Its slope is {eq}\displaystyle \frac{1}{2 } {/eq} and it contains the point {eq}(1, 1) {/eq} so its equation is given by:

{eq}\begin{align*} y - f(a) & = f'(a)(x-a) \\ y - 1 & =\frac{1}{2 }(x-1) \\ \implies y& = \frac{x}{2}+ \frac{1}{2}\\ \end{align*} {/eq}


Learn more about this topic:

Slopes of Tangent & Secant Lines

from TExES Mathematics 4-8 Exam (115): Study Guide & Review

Chapter 18 / Lesson 6
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