<span>h<span>(t)</span>=<span>t<span>34</span></span>−3<span>t<span>14</span></span></span>
Note that the domain of h is <span>[0,∞]</span>.
By differentiating,
<span>h'<span>(t)</span>=<span>34</span><span>t<span>−<span>14</span></span></span>−<span>34</span><span>t<span>−<span>34</span></span></span></span>
by factoring out <span>34</span>,
<span>=<span>34</span><span>(<span>1<span>t<span>14</span></span></span>−<span>1<span>t<span>34</span></span></span>)</span></span>
by finding the common denominator,
<span>=<span>34</span><span><span><span>t<span>12</span></span>−1</span><span>t<span>34</span></span></span>=0</span>
<span>⇒<span>t<span>12</span></span>=1⇒t=1</span>
Since <span>h'<span>(0)</span></span> is undefined, <span>t=0</span> is also a critical number.
Hence, the critical numbers are <span>t=0,1</span>.
I hope that this was helpful.
Answer:
1. 144 2. 16 3. 1 4. 3x-6
Step-by-step explanation:
So think of this as a function in a function. So you work from the inside to the outside. So for problem 1, we start with f(4)) [you read it "f of 4"] so what is the solution when x = 4, since f(x) means the function of x so f(4) means 'the function of 4' inside f(x).
Since f(x) = 3x then f(4) = 3(4) [notice how you substitute the 4 everywhere you see a letter x]
so f(4) = 12, now you work the next part h(f(4)) since f(4)=12 then h(12)
So take the h(x) function which is h(x) = 
then h(12) = 
so h(12) = 144
Answer:
Step-by-step explanation:9
