Question

Find the difference quotient mula1" title="\frac{f(x + h) - f(x)}{h}" alt="\frac{f(x + h) - f(x)}{h}" align="absmiddle" class="latex-formula"> for the following function. $f(x) = -x^2 -3x + 1$

Answer 1

Answer:

$-2x - h - 3$

Step-by-step explanation:

Step 1: Define

Difference Quotient: $\frac{f(x+h)-f(x)}{h}$

f(x) = -x² - 3x + 1

f(x + h) means that x = (x + h)

f(x) is just the normal function

Step 2: Find difference quotient

1. Substitute: $\frac{[-(x+h)^2-3(x+h)+1]-(-x^2-3x+1)}{h}$
2. Expand and Distribute: $\frac{[-(x^2+2hx+h^2)-3x-3h+1]+x^2+3x-1}{h}$
3. Distribute: $\frac{-x^2-2hx-h^2-3x-3h+1+x^2+3x-1}{h}$
4. Combine like terms: $\frac{-2hx-h^2-3h}{h}$
5. Factor out h: $\frac{h(-2x-h-3)}{h}$
6. Simplify: $-2x - h - 3$

Answer 2

Hey there!

We know that formula for the difference quotient is [ $\frac{f(x+h)-f(x)}{h}$ ]

We need to find f(x + h), so use (x + h) instead of x.

f(x + h) → -(x + h)² because we are given [ -x² ]

-3(h + x) + 1 because we are given [ -3x + 1 ]

Put what we found into the formula and simplify.

$\frac{(-(h+x)^2-3(h+x)+1-(x^2-3x+1)}{h}$

$-h-2x-3$

Therefore, the difference quotient for the given function is [ -h - 2x - 3 ]

Best of Luck!