Question

let ( f(x)=x^{3}-14x ). calculate the difference quotient ( \frac{f(3 + h)-f(3)}{h} ) for ( h = 1, h = 0.1, h=-0.1, h=-1 ). if someone now told you that the derivative (slope of the tangent line to the graph) of ( f(x) ) at ( x = 3 ) was an integer, what would you expect it to be? note: you can earn partial credit on this problem. 1 of the questions remains unanswered. at least one of the answers above is not correct.

Explanation:

Step1: Recall the difference - quotient formula

The difference - quotient formula is $\frac{f(x + h)-f(x)}{h}$. Here $x = 3$ and $f(x)=x^{3}-14x$. So $f(3)=3^{3}-14\times3=27 - 42=-15$. And $f(3 + h)=(3 + h)^{3}-14(3 + h)$.
Expand $(3 + h)^{3}$ using the formula $(a + b)^{3}=a^{3}+3a^{2}b + 3ab^{2}+b^{3}$, so $(3 + h)^{3}=27+27h + 9h^{2}+h^{3}$. Then $f(3 + h)=27+27h + 9h^{2}+h^{3}-42 - 14h=h^{3}+9h^{2}+13h - 15$.

Step2: Calculate the difference - quotient

$\frac{f(3 + h)-f(3)}{h}=\frac{(h^{3}+9h^{2}+13h - 15)-(-15)}{h}=\frac{h^{3}+9h^{2}+13h}{h}=h^{2}+9h + 13$.

Step3: Analyze the derivative as a limit

The derivative $f^\prime(3)=\lim_{h
ightarrow0}\frac{f(3 + h)-f(3)}{h}$. As $h$ approaches $0$, the value of the difference - quotient approaches the value of the derivative.
We can also find the derivative using the power rule. If $f(x)=x^{3}-14x$, then $f^\prime(x)=3x^{2}-14$. Substitute $x = 3$ into $f^\prime(x)$, we get $f^\prime(3)=3\times3^{2}-14=3\times9 - 14=27 - 14 = 13$.

Answer:

13