Local maximums and minimums always occur at values of t where the first derivative is zero (or undefined). So, the next step after you differentiate h(t) is to set that derivative equal to zero and solve for t. Use your technology to solve that equation.Okay. I’ll [find the first derivative of function h]. I wouldn’t know where to head after that … technology use is allowed
Sorry but that is not 100% true on a closed interval, as we have in this problem. Local max and mins can occur at the endpoints and the derivative may not be 0 there..Local maximums and minimums always occur at values of t where the first derivative is zero (or undefined). So, the next step after you differentiate h(t) is to set that derivative equal to zero and solve for t. Use your technology to solve that equation.
h'(t) = 0
There are two Real solutions; one of them is the t-value at the local maximum, and the other is the t-value at the local minimum.
You can use the second-derivative test, to confirm which is which. Has your class covered how that test works?
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Oh my. Yes, you're spot on, of course -- thank you for posting that. (A rookie mistake: peeking first at the graph of h versus thinking in general.)… on a closed interval … max and mins can occur at the endpoints ….
I did not hear you mention anything about corner time.Oh my. Yes, you're spot on, of course -- thank you for posting that. (A rookie mistake: peeking first at the graph of h versus thinking in general.)
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I don't know how to calculate the time.I did not hear you mention anything about corner time.