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Update linearApproximation1.tex #974

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2 changes: 1 addition & 1 deletion linearApproximation/exercises/linearApproximation1.tex
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Expand Up @@ -89,7 +89,7 @@
\begin {hint}
Recall: $f''(x)=-\frac{1}{4\cdot x^{\answer{\frac{3}{2}}}}$.

Since $f''(x)<0$ on $(0,\infty)$, the graph of $f$ is concave down. Therefore, the tangent line lies above the graph of $f$, except at $x=9$, where $f(9)=L(9)$.
Since $f''(x)<0$ on $\left(9,\answer{9.5}\right)$, the graph of $f$ is concave down. Therefore, the tangent line lies above the graph of $f$, except at $x=9$, where $f(9)=L(9)$.


This means that $L(x)\ge f(x)$. So, $L(9.5)\ge f(9.5)$.
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