Question

The twice-differentiable function f is defined for all real numbers and satisfies the following conditions:
f(0)=2, f'(0)=-4, and f''(0)=3

Answer 1

a.) 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. Show the work that leads to your answers.
b.) The function h is given by h(x)-cos(kx) f(x) for all real numbers, where k is a constant. Find h'(x) and write an equation for the line tangent to the graph of h at x=0

Answer 2

\[g(x)=e^{ax}+f(x)\\
\implies g'(x)=ae^{ax}+f'(x)\\
g''(x)=a^2e^{ax}+f''(x)\]

Answer 3

how would i find g'(0) and g''(0)

Answer 4

in \[g'(x)=ae^{ax}+f'(x)\] replace x by zero, and use the given information on f.

Answer 5

\[g'(0)\] means \[g'(x)\text{ when } x=0\]

Answer 6

\[g'(0)=a \times e ^{a \times 0} - 4\]

Answer 7

is that right?

Answer 8

@agent0smith

Answer 9

@RadEn

Answer 10

yeah, u were right. just simplify that
g′(0) = a * e^(a*0) − 4
g′(0) = a * e^0 − 4
g′(0) = a * 1 − 4
g′(0) = a − 4

Answer 11

what about g"(0), did u get it ?

Answer 12

g''(0)=a^2 + 3

Answer 13

correct too

Answer 14

how about for part b.)

Answer 15

can u rewrite that function, please

Answer 16

\[h(x)=\cos(kx)f(x)\]

Answer 17

nah, that's good :)

Answer 18

to get h' u have to use the product rule :
if given h = f * g then h' = gf' +fg'

Answer 19

is h'(x)=4cos(kx) + 4sin(kx)

Answer 20

h(x)=cos(kx)f(x)
so, h'(x) = f(x)(cos(kx))' + cos(kx)f '(x)
h'(x) = k f(x)(-sin(kx)) + cos(kx)f '(x)
h'(x) = -k f(x)sin(kx) + cos(kx)f '(x)

i dont know why there is 4 like yours

Answer 21

because the equation gives us f(0), f'(0), and f''(0)

Answer 22

Oh question b still be related with a, sorry im forgot

Answer 23

so was i correct? :P

Answer 24

h'(x) = -k f(x)sin(kx) + cos(kx)f '(x)
h'(0) = -k f(0)sin(k(0)) + cos(k(0))f '(0)
h'(0) = -k (2)sin(0) + cos(0)f '(0)
h'(0) = -k (2)(0) + 1(-4)
h'(0) = 0 - 4
h'(0) = - 4

Answer 25

woppps... missunderstanding :)

Answer 26

hup, canceled.. i have right
we just have h'(x) = -k f(x)sin(kx) + cos(kx)f '(x) only

then to get slope (m), we have to find the value of h'(0) and
i have done it above :P

Answer 27

so f'(0)=-4 is correct?

Answer 28

yes, h'(0) = -4 not f'(0)=-4

Answer 29

so how do i write an equation for the line tangent to the graph of h at x=0

Answer 30

there area 2 element to forms an equation of tangennt line are
the slope (we got) and the other is a point slope of that curve
in this case given a point slope x=0 (it just absis) then we have to determinen its ordinat point is y

Answer 31

to get the value of y, just subtitute x=0 to h(x), in other words find h(0)

Answer 32

h(x)=cos(kx)f(x)
y = h(0) = cos(k(0))f(0) = cos(0)f(0) = 1 * 2 = 2

Answer 33

so, the point slope at point (0,2)

Answer 34

y=-4x+4

Answer 35

now, use the formula straigh line equation :

with m is slope and (x1,y1) is the point slope