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Question: How do you find and classify all the critical points and then use the second derivative to check you...

How do you find and classify all the critical points and then use the second derivative to check your results given h(t)=−4.9t2+39.2t+2h\left( t \right) = - 4.9{t^2} + 39.2t + 2 ?

Explanation

Solution

Hint : The second derivative of the given function is used to find out the critical points i.e., Maxima and Minima where the first derivative does not give the same for the given function. If a function has a critical point for which h′(t)=0h'\left( t \right) = 0 and the second derivative is positive at this point, then h has a local minimum here. If, however, the function has a critical point for which h′(t)=0h'\left( t \right) = 0 and the second derivative is negative at this point, then h has a local maximum here.

Complete step by step solution:
Given,
h(t)=−4.9t2+39.2t+2h\left( t \right) = - 4.9{t^2} + 39.2t + 2
You first need to find the critical points where h′(t)=0h'\left( t \right) = 0 , and then check the sign of the second derivative in those points.
Let's calculate first the first derivative i.e., h′(t)h'\left( t \right)
h(t)=−4.9t2+39.2t+2h\left( t \right) = - 4.9{t^2} + 39.2t + 2
We know that, ddt(t2)=2t\dfrac{d}{{dt}}\left( {{t^2}} \right) = 2t and ddt(t)=1\dfrac{d}{{dt}}\left( t \right) = 1 hence applying this we get:
⇒h′(t)=−4.9⋅2⋅t+39.2\Rightarrow h'\left( t \right) = - 4.9 \cdot 2 \cdot t + 39.2
Hence, h′(t)=0h'\left( t \right) = 0 means,
⇒h′(t)=−9.8t+39.2=0\Rightarrow h'\left( t \right) = - 9.8t + 39.2 = 0 , and thus, the only critical point is t=4t = 4 .
Now let us calculate the second derivative in t=4t = 4 . But the second derivative is −9.8, so in particular the second derivative in t=4t = 4 is −9.8.
As the second derivative is negative in the critical point, the function has a local maximum at that point.

Note : We must note that, if t is a critical point of h(t)h\left( t \right) and the second derivative of h(t)h\left( t \right) is positive, then x is a local minimum of h(t)h\left( t \right) . Likewise, if t is a critical point of h(t)h\left( t \right) and the second derivative of h(t)h\left( t \right) is negative, then the slope of the graph of the function is zero at that point, i.e., if h′(t)<0h'\left( t \right) < 0 the function has a maximum at that point and h′(t)>0h'\left( t \right) > 0 the function has a minimum at that point.