# differntiable functions

• Oct 24th 2007, 04:56 AM
DINOCALC09
differntiable functions
The twice-differentiable function f is defined for all real numbers and satisies the following conditions:

f(0) = 2
f'(0) = -4
f''(0) = 3

The function g is given by g(x) = e^(ax) + f(x) for all real numbers, where a is a constant. Find g'(0) and g''(0) in terms of a.
• Oct 24th 2007, 07:04 AM
earboth
Quote:

Originally Posted by DINOCALC09
a) The twice-differentiable function f is defined for all real numbers and satisies the following conditions:

f(0) = 2
f'(0) = -4
f''(0) = 3

b) The function g is given by g(x) = e^(ax) + f(x) for all real numbers, where a is a constant. Find g'(0) and g''(0) in terms of a.

Hello,

to a):

Assume that f''(x) = 3 then

$f'(x) = \int(f''(x))dx = 3x+C$. With f'(0) = -4 you'll get C = -4

Thus f'(x) = 3x-4

Then $f(x) = \int(f'(x)dx = \frac32 x^2-4x+C$ . With f(0) = 2 you'll get C = 2 and therefore the function is:

$f(x) = \frac32 x^2 -4x + 2$

to b)
I assume that you have to use the function f from a). Thus

$g(x) = e^{ax}+f(x)~\implies~g'(x)=a \cdot e^{ax}+f'(x)~\implies~g''(x)=a^2 \cdot e^{ax}+f''(x)$ Plug in x = 0 (Remember $e^0 = 1$) and the values for f'(0) and f''(0).

$g'(0) = a \cdot 1 -4 = a-4$

$g''(x) = a^2+3$