Solve the problem.
-Find an equation of the tangent to the curve $f ( x ) = \sqrt { x + 1 }$ that has slope $\frac { 1 } { 4 }$.
A) $y = - \frac { 1 } { 4 } x + \frac { 5 } { 4 }$
B) $y = \frac { 1 } { 4 } x + \frac { 5 } { 4 }$
C) $y = \frac { 1 } { 4 } x - \frac { 5 } { 4 }$
D) $y = \frac { 1 } { 4 } x$

Question

Quizplus · Accepted Answer

First, we find the derivative of the curve:
$$f'(x) = \frac{1}{2\sqrt{x+1}}$$
Next, we set $f'(x)$ equal to the given slope of $\frac{1}{4}$ and solve for $x$:
$$\frac{1}{2\sqrt{x+1}} = \frac{1}{4} \implies \sqrt{x+1} = 2 \implies x=3$$
Now, we use the point-slope form of the equation of a line with point $(3, 2)$ and slope $\frac{1}{4}$:
$$y-2 = \frac{1}{4}(x-3) \implies y = \frac{1}{4}x + \frac{5}{4}$$
Therefore, the equation of the tangent line is $y = \frac{1}{4}x + \frac{5}{4}$, which corresponds to choice (B).

Solve the Problem f(x)=x+1f ( x ) = \sqrt { x + 1 }f(x)=x+1​

Solve the Problem $f ( x ) = \sqrt { x + 1 }$