hello all
if u can plz help me with this homework i attached it here
thx 4 all
$\displaystyle f(x) = 6x - e^{1 - 2x}$
a)
$\displaystyle A = \int_1^5 (6x - e^{1 - 2x})dx$
and
$\displaystyle W = \frac{1}{4 - 1} \int_1^4 (6x - e^{1 - 2x})dx$
To do either of these we need
$\displaystyle \int (6x - e^{1 - 2x})dx$
Let's rewrite this a little:
$\displaystyle = \int (6x - e^1 \cdot e^{-2x})dx$
$\displaystyle = \int 6x~dx - e \int e^{-2x}~dx$
The first integral is elementary, for the second one use the substitution u = -2x and the fact that $\displaystyle \int e^u ~ du = e^u + C$.
b)
$\displaystyle f^{\prime}(x) = 6 - e^{1 - 2x} \cdot -2 = 6 + 2e^{1 - 2x}$
$\displaystyle f^{\prime \prime}(x) = 2e^{1 - 2x} \cdot -2 = -4e^{1 - 2x}$
So just plug these into the expression $\displaystyle f^{\prime \prime} + 2f^{\prime} - 12$ and show that it has a value of 0.
c)
Concavity is determined by the second derivative. So to get the inflection points solve $\displaystyle f^{\prime \prime}(x) = 0$ for x. (Hint: $\displaystyle e^{x} \geq 0$ for all x, so are there any values of x such that the second derivative is 0?)
To determine the concavity of the function, if $\displaystyle f^{\prime \prime}(x) > 0$ the function is concave up, if $\displaystyle f^{\prime \prime}(x) < 0$ the function is concave down.
-Dan