Problemas

Let f(x)=x^2+9x+20 and g(x)=x^3-4x Find (fg)(x) and ((f)/(g))(x) (fg)(x)=square ((f)/(g))(x)=square State the domain of each. Domain of (fg)(x) square Domain of ((f)/(g))(x) = square : all real numbers xvert xneq 0 xvert xneq -2,0,2 Evaluate the following. (fg)(-1)=square ((f)/(g))(-1)= square xvert xneq 0,2
Solución
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Pedroprofessionell · Tutor durante 6 años
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\((fg)(x) = (x^2 + 9x + 20) \cdot (x^3 - 4x) = x^5 + 5x^4 - 16x^2 - 80x\)<br /><br />\((f/g)(x) = (x^2 + 9x + 20) / (x^3 - 4x) = (x + 4) / (x - 2)\)<br /><br />The domain of \((fg)(x)\) is all real numbers.<br /><br />The domain of \((f/g)(x)\) is all real numbers except \(x = 0\) and \(x = 2\).<br /><br />\((fg)(-1) = (-1)^5 + 5(-1)^4 - 16(-1)^2 - 80(-1) = -1 + 5 - 16 + 80 = 68\)<br /><br />\((f/g)(-1) = (-1 + 4) / (-1 - 2) = 3 / -3 = -1\)
Explicar
## Step 1<br />The problem involves the operations of function multiplication and division. The functions given are \(f(x) = x^2 + 9x + 20\) and \(g(x) = x^3 - 4x\).<br /><br />## Step 2<br />To find the product of the two functions, we multiply them together. This is represented as \((fg)(x) = f(x) \cdot g(x)\).<br /><br />## Step 3<br />Similarly, to find the quotient of the two functions, we divide \(f(x)\) by \(g(x)\). This is represented as \((f/g)(x) = f(x) / g(x)\).<br /><br />## Step 4<br />The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the product and quotient of two functions, the domain is the set of all real numbers except for those that make the denominator zero.<br /><br />## Step 5<br />To evaluate \((fg)(-1)\) and \((f/g)(-1)\), we substitute \(x = -1\) into the expressions for \((fg)(x)\) and \((f/g)(x)\).
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