anonymous
  • anonymous
find the domain and range of the given function y(t)=3t^2-2t+1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Unless a domain is given explicitly with a function, we assume the domain is the set of all t (here) for which the function is defined (i.e. for which the function 'works'). In this case, you can take any value for t, plug it in, and the function will work. Your domain is therefore,\[D=\left\{ t |t \in \mathbb{R} \right\}\] (read, "The set of all t, such that t is an element of the real numbers). The range is the set of all y-values that your function will take. If you plot this function, you'll see that it's a parabola. This parabola will have a minimum value at some point, and then extend upwards to infinity from there. Your range, therefore, won't be all the real numbers, unlike your domain. To find the minimum, you can either complete the square and find the t-value that makes the function minimal, or use calculus to find that the point at which f(t) is minimal occurs at \[t=\frac{1}{3}\]and so\[f(\frac{1}{2})=3(\frac{1}{2})^2-2(\frac{1}{2})+1=\frac{3}{4}\]The function minimum is therefore\[\frac{3}{4}\]The function can take values above this value, and at this value, but not below. So your function's range is\[R=\left\{ f(t)|f(t) \in \mathbb{R}, f(t) \gt \frac{3}{4} \right\}\]which is read, "The range of f is the set of all values f(t) such that f(t) is real and f(t) is greater than 3/4." If you need clarification on anything, please let me know :)
anonymous
  • anonymous
thanx but i didnt quite get it though

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