Find values of a and b so that the function has a local maximum at the point (12, 36).
f(x) = axe^(bx)
I really am just confused on how to do this.
Thanks in advance,
Tyler
First, derive the function:
$\displaystyle f'(x) = ae^{bx} + abxe^{bx} = e^{bx}(a + abx)$
Now, for stationary points, the derivative must equal 0.
So $\displaystyle 0 = e^{bx}(a+ abx)$
Since $\displaystyle e^{bx}$ can never equal 0, we must have $\displaystyle (a + abx) = 0 \therefore x = \frac{-1}{b}$
For the stationary point to be at 12, we must have $\displaystyle 12 = \frac{-1}{b} \therefore b = \frac{-1}{12}$
Going back to the original function, we must now have:
$\displaystyle 36 = 12ae^{-1}$
$\displaystyle 3 = \frac{a}{e} \therefore a = 3e$