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Mathematics 13 Online
OpenStudy (anonymous):

The equation of the tangent line to f(x) = √x at x = 49 is y = Using this, we find our approximation for √{49.2} is..?

zepdrix (zepdrix):

Did you find the equation for the tangent line yet?

OpenStudy (anonymous):

no i don't know how to do this question

zepdrix (zepdrix):

So a line is of the form:\[\Large\bf\sf y=mx+b\]The line tangent to our curve will have a special `slope`:\[\Large\bf\sf m=f'(49)\]The derivative evaluated at 49 gives us the slope of the line tangent to our curve at that point.

zepdrix (zepdrix):

\[\Large\bf\sf f(x)=\sqrt x, \qquad\qquad f'(x)=?\]Do you understand how to find this derivative?

OpenStudy (anonymous):

is it 7

zepdrix (zepdrix):

You evaluated the `function` at x=49,\[\Large\bf\sf f(49)=\sqrt{49}\]\[\Large\bf\sf f(49)=7\]That gives us an ordered pair, \(\Large\bf\sf (49,\;7)\) That's good, we'll need that information later. Do you understand how to find the derivative of the function that was given?

OpenStudy (anonymous):

nope sorry..

zepdrix (zepdrix):

You can rewrite \(\Large\bf\sf \sqrt x\) as \(\Large\bf\sf x^{1/2}\) and then just apply the power rule from there. But the square root shows up so often that it's worth memorizing, so try to remember this one,\[\Large\bf\sf f(x)=\sqrt x,\qquad\qquad f'(x)=\frac{1}{2\sqrt x}\]Derivative of square root is one over 2 square roots.

zepdrix (zepdrix):

Evaluating the derivative at 49 gives us the `slope` of our tangent line: \[\Large\bf\sf f'(49)=\frac{1}{2\sqrt{49}}\]

zepdrix (zepdrix):

So we have this tangent line,\[\Large\bf\sf y=mx+b\]We get the slope, `m`, from the derivative evaluated at x=49,\[\Large\bf\sf y=\frac{1}{14}x+b\]To find the y-intercept, `b`, we need to plug in the ordered pair that we came up with earlier.

OpenStudy (anonymous):

ooooh so for b i would plug in 7?

zepdrix (zepdrix):

We don't plug anything in directly for b, we're trying to solve for b. We have an ordered pair that this line passes through,\[\Large\bf\sf (x,\;y)=(49,\;7)\]We plug those in for x and y and proceed to solve for b,\[\Large\bf\sf 7=\frac{1}{14}\cdot 49+b\]

zepdrix (zepdrix):

So,\[\Large\bf\sf 7=\frac{49}{14}+b\] Let's subtract 49/14 from each side,\[\Large\bf\sf 7-\frac{49}{14}=b\]Which tells us that our b is 3.5 I think. My math look right?

zepdrix (zepdrix):

Here is a graph to illustrate what we're doing.

OpenStudy (anonymous):

i typed in 3.5 as my answer for y and it says it's wrong

zepdrix (zepdrix):

|dw:1395882034058:dw|

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