Find the equation of the tangent line to f at x=a. Given the function f(x)=sqroot(x+3), and a=1

Question

Slope of the tangent line is 1/4

Answer 1

If f(x) = sqrt(x+3), then we can say that f(x) = (x+3)^(1/2) and we can apply a power rule to get that f'(x), the slope of the curve at any value of x is 1/2(x+3)^(-1/2).

Note that when we looked inside our parenthesis for a potential chain rule factor, we found none because (x-3)' is just 1.

Note also that I said "at any value of x." When we differentiated, we found that f'(x) is not defined when x+3 is not positive because we cannot take the square root of a negative number and we can't divide by zero.

But, we are concerned about when x = 1. When x = 1, f'(x) = 1/[2*sqrt(4)] = 1/4.

We know that the function passes through (1,2). So we now have a slope and a point and we can create another hypothetical point on the line, (x,y) and we can say that (y-2)/(x-1) = 1/4, and that's an equation of the required line since the slopes between any two points on the (straight) tangent line are equal everywhere.

Answer 2

At x=1 the function passes through (1,2).

Yes, the slope of the tangent is 1/4.

The equation of the tangent line is 1/4 = (y-2)/(x-1) and

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