0. Let f (x) = x^4
:
(a) Find a linear approximation of f (x) at a = 1
(b)If dx = x = 0.25; what are dy and deltay?

Question

0. Let f (x) = x^4
:
(a) Find a linear approximation of f (x) at a = 1
(b)If dx = x = 0.25; what are dy and deltay?

anonymous · Answer

for a i got 4x-3 but not sure if its suppose to be an equation?

anonymous · Answer

and b i know dy=1 but will delta x also equal 1?

phi · Answer

f(x) = x^4
Yes, part a looks good.

at x= 1, f(1) = 1^4 =1,   and (1,1) is on the curve
you want the equation of a line through that point, with a slope df/dx  evalutated at x=1

the equation of the line using point-slope form is
y - y0 = m(x- x0)

with m= df/dx, the equation through (1,1) is
y - 1 = m(x - 1)
or
y = mx -m+1

df/dx = 4x^3 .   At x=1  df/dx = 4
y = 4x  -4 +1   or
y = 4x -3

Here is a plot:

anonymous · Answer

okay thanks..DO you know how to solve part B?

phi · Answer

I think 
$$ \Delta y = \frac{dy}{dx} \Delta x $$

anonymous · Answer

do you think i could solve delta y by f(1.25)-f(1)  or would i have to use an equation?

phi · Answer

where did you get 1.25?  (b) says x = 0.25  (and Δx= 0.25)

phi · Answer

If x= 1/4  and Δx= 1/4  I would do this
dy/dx = 4 x^3 evaluated at x= 1/4-->  dy/dx = 1/16
Δy = 1/16* Δx = 1/16 * 1/4 = 1/64

anonymous · Answer

okay i see read the problem wrong thanks