Science:Math Exam Resources/Courses/MATH102/December 2012/Question C 03 (b)/Solution 1

Using the linear approximation formula

$f(x)\approx f^{\prime }(a)(x-a)+f(a)$

where our function is $f(x)=\sqrt{x}$ and our value of a will be a number near 48. Choosing a = 49 is sensible because you can take the square root easily. We can either calculate $f^{\prime }(x)$ again or note that we already found it in question C2 as $f^{\prime }(x)=1/(2\sqrt{x})$. Plugging all this in gives:

$f(x)\approx \frac{1}{2\sqrt{49}}(x-49)+\sqrt{49}=\frac{1}{14}(x-49)+7$

To approximate $b=\sqrt{48}$, we say

$\sqrt{48}=f(48)\approx \frac{1}{14}(48-49)+7=\frac{-1}{14}+7=6\frac{13}{14}$

Whether the linear approximation is an over or under estimate depends on the concavity of the function - if $f(x)$ is concave up, the tangent line approximation lies below the graph and is an under-estimate; the opposite is true if $f(x)$ is concave down. Because $f(x)=\sqrt{x}$ is concave down, our estimate in this problem is an over-estimate.