91Ó°ÊÓ

We are given the function: \(f(x) = x^3 - 2x\). To find \(f(x+h)\) and \(f(a)\), we substitute \((x+h)\) and \(a\) into the expression: \(f(x+h) = (x+h)^3 - 2(x+h)\) \(f(a) = a^3 - 2a\)

Now, we will plug in our expressions for \(f(x+h)\) and \(f(x)\) into the first difference quotient: \(\frac{f(x+h)-f(x)}{h} = \frac{((x+h)^3 - 2(x+h)) - (x^3 - 2x)}{h}\) To simplify, expand \((x+h)^3\) using the binomial theorem and then combine like terms: \(\frac{f(x+h)-f(x)}{h} = \frac{(x^3 + 3x^2h + 3xh^2 + h^3 - 2x - 2h) - (x^3 - 2x)}{h}\) \(\frac{f(x+h)-f(x)}{h} = \frac{3x^2h + 3xh^2 + h^3 - 2h}{h}\) Now, we can factor out an \(h\) from every term in the numerator: \(\frac{f(x+h)-f(x)}{h} = \frac{h(3x^2 + 3xh + h^2 - 2)}{h}\) Finally, cancel out the \(h\) term: \(\frac{f(x+h)-f(x)}{h} = 3x^2 + 3xh + h^2 - 2\)

Next, we will plug in our expressions for \(f(x)\) and \(f(a)\) into the second difference quotient: \(\frac{f(x)-f(a)}{x-a} = \frac{(x^3 - 2x) - (a^3 - 2a)}{x-a}\) To simplify, combine like terms: \(\frac{f(x)-f(a)}{x-a} = \frac{x^3 - a^3 - 2x + 2a}{x-a}\) Now, we can use the difference of cubes formula for the terms \(x^3 - a^3\): \(x^3 - a^3 = (x-a)(x^2 + ax + a^2)\) Substitute this expression back into our difference quotient: \(\frac{f(x)-f(a)}{x-a} = \frac{(x-a)(x^2 + ax + a^2) - 2x + 2a}{x-a}\) Lastly, we can cancel out the \((x-a)\) term: \(\frac{f(x)-f(a)}{x-a} = x^2 + ax + a^2 - 2x + 2a\) So the simplified difference quotients are: \(\frac{f(x+h)-f(x)}{h} = 3x^2 + 3xh + h^2 - 2\) \(\frac{f(x)-f(a)}{x-a} = x^2 + ax + a^2 - 2x + 2a\)