Problem 32 - Precalculus Essentials

Question

Evaluate each function at the given values of the independent variable and simplify. \(h(x)=x^{3}-x+1\) a. \(h(3)\) b. \(h(-2)\) c. \(h(-x)\) d. \(h(3 a)\)

Step-by-Step Solution

Answer

a. \(h(3) = 25\), b. \(h(-2) = -5\), c. \(h(-x) = -x^{3} + x + 1\), d. \(h(3a) = 27a^{3} - 3a + 1\)

Step 1: Evaluate \(h(3)\)

Substitute \(x = 3\) into the function \(h(x)\) to get \(h(3) = 3^{3} - 3 + 1\). Simplify the expression to get \(h(3) = 27 - 3 + 1 = 25\).

Step 2: Evaluate \(h(-2)\)

This time substitute \(x = -2\) into the function \(h(x)\) to get \(h(-2) = (-2)^{3} - (-2) + 1\). Simplify this expression to get \(h(-2) = -8 + 2 + 1 = -5\).

Step 3: Evaluate \(h(-x)\)

Substitute \(x = -x\) into the function \(h(x)\) giving us \(h(-x) = (-x)^{3} - (-x) + 1\). Here, we have a simpler formula \(h(-x) = -x^{3} + x + 1\)

Step 4: Evaluate \(h(3a)\)

Finally, substitute \(x = 3a\) into the function \(h(x)\) resulting in \(h(3a) = (3a)^{3} - 3a + 1 = 27a^{3} - 3a + 1\).