Course:MATH102/Question Challenge/2008 December Q1.2

	Questions? Click here to add or read comments for this problem
Please rate how easy you found this problem: {{#w4grb_rate:}} Hard                                              Easy

Question

The functions $f(x)=x^{2}$ and $g(x)=x^{3}$ are equal at $x=0$ and at $x=1$. Between $x=0$ and at $x=1$, for what value of $x$ are their graphs furthest apart?

(a) $x=1/2$

(b) $x=2/3$

(c) $x=1/3$

(d) $x=1/4$

(e) $x=3/4$

Hints

Hint
add your hints here

Solutions

Solution

$f(x)=x^{2}$ $g(x)=x^{3}$ Within the domain $0\leq x\leq 1$, lower power functions dominate higher power functions. Therefore we will take the derivative of $f(x)-g(x)$ and solve for the local maximum: $h(x)=f(x)-g(x)$ $h(x)=x^{2}-x^{3}$ ${\frac {dh}{dx}}=2x-3x^{2}$ $2x=3x^{2}$ $2=3x$ $x=2/3$ To prove that this is a local maximum and not a local minimum we will take the second derivative: ${\frac {dh}{dx}}=2x-3x^{2}$ ${\frac {d^{2}h}{dx}}=2-6x$ Substituting in for $x=2/3$: ${\frac {d^{2}h}{dx}}=2-6(2/3)$ ${\frac {d^{2}h}{dx}}=2-4$ ${\frac {d^{2}h}{dx}}=-2$ When the second derivative is negative and the first derivative is equal to zero we have found a local maximum. The difference in the values between $f(x)$ and $g(x)$ is greatest when $x=2/3$