Justify that you have found the requested point by analyzing an appropriate derivative. x=t+1 y=t2+t −2≤t≤2 Lowet point.

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Justify that you have found the requested point by analyzing an appropriate derivative.  x=t+1  y=t2+t  −2≤t≤2  Lowet point.

Jenny Sheppard · Accepted Answer

Step 1 Since we need to find the lowest point of the curve, we need to find the minimum of the function y=t2+t, where −2≤t≤2 We know that function y has a relative minimum at point t=t0 if e have that dydt(t0)=0 dydt&amp;lt;0 for t&amp;lt;t0, and dydt&amp;gt;0 for t&amp;gt;t0. We also know that if function y has one relative minimum at point t=t0, then function y has the minimum at point t=t0 Using the previous result, we have that dydt=2t+1⇒dydt(−0.5)=0 Using the previous results, we also have that dydt&amp;lt;0 for −2≤t&amp;lt;−0.5 and dydt&amp;gt;0 for -0.5 According to the previous results, we can conclude that the function y=t2+t, where −2≤t≤2, has the minimum at point t=−0.5. Since in this exercise we have that x=t+1, where −2≤t≤2, then we have that x(−0.5)=−0.5+1=0.5 y(−0.5)=(−0.5)2+(−0.5) =0.25−0.5=−0.25 Which means that the lowest point of curve is equal to (0.5, −0.25)

Justify that you have found the requested point by analyzing

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