91Ó°ÊÓ

Before sketching out the graph, we need to understand what each of the four conditions says: 1. The domain of \(f\) is \([-2, 4]\): This means that the function is defined only between \(-2\) and \(4\), inclusive. 2. The range of \(f\) is \([-5, 6]\): This means that the output, or the \(y\)-values of the function, lie between \(-5\) and \(6\), inclusive. 3. \(f(-1)=f(3)\): This means that the function has the same output at both \(x=-1\) and \(x=3\). 4. \(f\left(\frac{1}{2}\right)=0\): This means that the function crosses the x-axis at \(x=\frac{1}{2}\).

Now let's try to construct a function that satisfies these conditions. We can consider a simple quadratic function, since it might have ample flexibility accomplish what we need. Let's suppose the function is: $$ f(x) = ax^2 + bx + c $$ Now, we will use the given conditions to find the values of \(a\), \(b\), and \(c\).

Substitute \(x = -1\) and \(x = 3\) into the quadratic function and equate the two expressions: $$ a(-1)^2 + b(-1) + c = a(3)^2 + b(3) + c $$ This simplifies to: $$ a - b + c = 9a + 3b + c $$ Which implies: $$ 8a + 4b = 0 $$ This means that \(a\) and \(b\) must be related, we can simplify further by dividing by 4: $$ 2a + b = 0 $$

Substitute \(x = \frac{1}{2}\) into the quadratic function and set it to zero: $$ a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + c = 0 $$ This simplifies to: $$ \frac{1}{4}a + \frac{1}{2}b + c = 0 $$

We will now choose values of \(a\), \(b\), and \(c\) that satisfy the conditions and simplifies our task. We can pick \(a=-1\), because by observing that with \(2a+b=0\), it implies \(b=2\). Finally, we should figure out what value of \(c\) can fit both the domain and range criteria. After trial-and-error, we can choose \(c=6\). By choosing these values, we obtain the function: $$ f(x) = -x^2 + 2x + 6 $$

Now that we have our function, let's make sure it fits the domain and range criteria. The domain of the function is automatically satisfied since \(f(x)\) is defined for all real numbers. To verify the range, we need to find the minimum value of the function. For a quadratic function, the vertex of the parabola occurs at \(x = -\frac{b}{2a}\). In our case, this would be at \(x = -\frac{2}{-2}=1\). Substituting, we get: $$ f(1) = -1^2 + 2(1) + 6 = -1 + 2 + 6 = 7 $$ Since the vertex is the maximum point and it lies within the desired range, we can conclude that this function satisfies all four conditions. To plot the graph of the function, simply sketch a parabola with the vertex at \((1, 7)\), crossing the x-axis at \(x=\frac{1}{2}\), and with equal \(y\)-values at \(x=-1\) and \(x=3\). Make sure to restrict the domain to \([-2,4]\) and the range to \([-5,6]\).