let (f(x)=x^{3}-14x). calculate the difference quotient (\frac{f(3 + h)-f(3)}{h}) for (h = 1), (h = 0.1), (h…

let (f(x)=x^{3}-14x). calculate the difference quotient (\frac{f(3 + h)-f(3)}{h}) for (h = 1), (h = 0.1), (h = 0.01), (h=-0.01), (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.

let (f(x)=x^{3}-14x). calculate the difference quotient (\frac{f(3 + h)-f(3)}{h}) for (h = 1), (h = 0.1), (h = 0.01), (h=-0.01), (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.

Answer

Explanation:

Step1: Recall the difference - quotient formula

The difference - quotient formula for a function $y = f(x)$ 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 $f(3 + h)=(3 + h)^{3}-14(3 + h)$.

Step2: Expand $(3 + h)^{3}-14(3 + h)$

Using the formula $(a + b)^{3}=a^{3}+3a^{2}b + 3ab^{2}+b^{3}$, we have $(3 + h)^{3}=27+27h + 9h^{2}+h^{3}$. Then $(3 + h)^{3}-14(3 + h)=27+27h + 9h^{2}+h^{3}-42-14h=h^{3}+9h^{2}+13h - 15$.

Step3: Calculate the difference - quotient

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

Step4: Evaluate for different values of $h$

When $h = 1$: Substitute $h = 1$ into $h^{2}+9h + 13$, we get $1^{2}+9\times1+13=1 + 9+13=23$. When $h = 0.1$: Substitute $h = 0.1$ into $h^{2}+9h + 13$, we get $(0.1)^{2}+9\times0.1+13=0.01 + 0.9+13=13.91$. When $h=-0.01$: Substitute $h=-0.01$ into $h^{2}+9h + 13$, we get $(-0.01)^{2}+9\times(-0.01)+13=0.0001-0.09 + 13=12.9101$. When $h=-1$: Substitute $h=-1$ into $h^{2}+9h + 13$, we get $(-1)^{2}+9\times(-1)+13=1-9 + 13=5$. As $h$ approaches $0$, the difference - quotient approaches the derivative of $f(x)$ at $x = 3$. The derivative of $f(x)=x^{3}-14x$ using the power rule $f^\prime(x)=3x^{2}-14$. When $x = 3$, $f^\prime(3)=3\times3^{2}-14=3\times9-14=27 - 14 = 13$.

Answer:

When $h = 1$, the difference - quotient is 23; when $h = 0.1$, it is 13.91; when $h=-0.01$, it is 12.9101; when $h=-1$, it is 5. We would expect the derivative (the value the difference - quotient approaches as $h\to0$) to be 13.