Question

let $f(x)=x^{3}-6x$. calculate the difference quotient $\frac{f(2 + h)-f(2)}{h}$ for $h = .1$ $h = .01$ $h =-.01$ $h =-.1$ if someone now told you that the derivative (slope of the tangent line to the graph) of $f(x)$ at $x = 2$ was an integer, what would you expect it to be?

Explanation:

Step1: Find \(f(2 + h)\) and \(f(2)\)

First, find \(f(2+h)=(2 + h)^3-6(2 + h)=8 + 12h+6h^{2}+h^{3}-12 - 6h=h^{3}+6h^{2}+6h - 4\). And \(f(2)=2^{3}-6\times2=8 - 12=-4\). Then \(\frac{f(2 + h)-f(2)}{h}=\frac{h^{3}+6h^{2}+6h - 4-(-4)}{h}=\frac{h^{3}+6h^{2}+6h}{h}=h^{2}+6h + 6\).

Step2: Calculate when \(h = 0.1\)

Substitute \(h = 0.1\) into \(h^{2}+6h + 6\). We get \((0.1)^{2}+6\times0.1+6=0.01 + 0.6+6=6.61\).

Step3: Calculate when \(h = 0.01\)

Substitute \(h = 0.01\) into \(h^{2}+6h + 6\). We get \((0.01)^{2}+6\times0.01+6=0.0001+0.06 + 6=6.0601\).

Step4: Calculate when \(h=-0.01\)

Substitute \(h=-0.01\) into \(h^{2}+6h + 6\). We get \((-0.01)^{2}+6\times(-0.01)+6=0.0001-0.06 + 6=5.9401\).

Step5: Calculate when \(h=-0.1\)

Substitute \(h=-0.1\) into \(h^{2}+6h + 6\). We get \((-0.1)^{2}+6\times(-0.1)+6=0.01-0.6 + 6=5.41\).

Step6: Estimate the derivative

As \(h\) approaches \(0\), the difference - quotient approaches the derivative. The derivative of \(y = f(x)=x^{3}-6x\) is \(y'=3x^{2}-6\). When \(x = 2\), \(y'=3\times2^{2}-6=12 - 6 = 6\).

Answer:

When \(h = 0.1\), the value is \(6.61\).
When \(h = 0.01\), the value is \(6.0601\).
When \(h=-0.01\), the value is \(5.9401\).
When \(h=-0.1\), the value is \(5.41\).
If the derivative at \(x = 2\) is an integer, it is \(6\).