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

Using the linear approximation formula

$f(x)\approx f'(a)(x-a)+f(a)$

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 $\displaystyle f'(x)$ again or note that we already found it in question C2 as $f'(x)=1/(2{\sqrt {x}})$. 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.