Math  /  Calculus

QuestionUse the graph below to find the largest value of δ>0\delta>0 such that for all x,f(x)L<εx,|f(x)-L|<\varepsilon whenever 0<xc<δ0<|x-c|<\delta. f(x)=3xc=1L=3ε=0.5\begin{aligned} f(x) & =\frac{3}{\sqrt{-x}} \\ c & =-1 \\ L & =3 \\ \varepsilon & =0.5 \end{aligned}
The largest value of δ\delta is \square (Simplify your answer.)

Studdy Solution
The center of the interval is c=1 c = -1 .
Calculate δ \delta as the distance from c c to the nearest endpoint of the interval: δ=min(1+3625,1+3649) \delta = \min\left( \left| -1 + \frac{36}{25} \right|, \left| -1 + \frac{36}{49} \right| \right)
Calculate: 1+3625=1125 \left| -1 + \frac{36}{25} \right| = \frac{11}{25} 1+3649=1349 \left| -1 + \frac{36}{49} \right| = \frac{13}{49}
Convert to decimals for comparison: 1125=0.44 \frac{11}{25} = 0.44 13490.2653 \frac{13}{49} \approx 0.2653
The largest δ \delta is the smaller value: δ=1349 \delta = \frac{13}{49}
The largest value of δ \delta is:
1349 \boxed{\frac{13}{49}}

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