Problem 86 - Precalculus Essentials

Question

Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=-3 x^{2}+x-1$$

Step-by-Step Solution

Answer

The simplified difference quotient for the given function is \( -6x - 3h + 1\)

Step 1: Substitute the function into the difference quotient

To determine the difference quotient, replace all instances of \(x\) in \(f(x)\) with \(x + h\). This gives us \(f(x + h) = -3(x + h)^{2} + (x + h) - 1\) then simplify this expression and the original function \(f(x)\) to give \( \frac{-3(x + h)^{2} + x + h - 1 - (-3x^{2} + x - 1)} {h}\)

Step 2: Simplify the numerator

Next, simplify the numerator by expanding and combining like terms, we can expand \( (x+h)^{2}\) to get \(x^{2} + 2hx + h^{2}\). So, the expanded equation becomes \( \frac{-3x^{2} - 6hx -3h^{2} + x + h -1 +3x^{2} - x +1}{h}\). Further simplify the equation by combining like terms to get \( \frac{- 6hx - 3h^{2} + h}{h}\)

Step 3: Cancel out the common factor

Finally, cancel out the common factor of \(h\) from the numerator and denominator. This leaves us with \( -6x - 3h + 1\)