Find the linearization L(x) of the function g(x) = xf(x^2) at x = 2 given the following information. f(2)=1 f'(2)=8 f(4)=4 f'(f)=-3
L(x)=
g'(x) = f(x^2) + xf'(x^2)(2x)
= f(x^2) + 2x^2 f'(x^2)
So, at x=2,
g'(2) = f(4) + 8f'(4) = 4+8(-3) = -20
g(2) = 2f(4) = 8
So, now we have a point and a slope:
L(x) = -20(x-2) + 8 = -20x+48
Oh, linearization, the art of making things simpler! Let's do this!
First, let's find the value of f(2) and f'(2), I hope they're not hiding from us! Ah, there they are, f(2) = 1 and f'(2) = 8.
Now, let's find the slope of the tangent line at x = 2. We can use the formula for linearization: L(x) = f(a) + f'(a)(x - a).
Oh, silly me! I almost forgot to mention! We need to find f(4) too, which is equal to 4. Don't worry, we're professional clown-bots, we always remember to include all the important details!
So, now we have all the ingredients we need. Let's put them together: L(x) = f(2) + f'(2)(x - 2).
Substituting in the values we found: L(x) = 1 + 8(x - 2).
And there you have it! The linearization L(x) of the function g(x) = xf(x^2) at x = 2 is L(x) = 1 + 8(x - 2).
To find the linearization of the function g(x) = xf(x^2) at x = 2, we need to use the concept of the linear approximation. The linearization can be approximated using the formula:
L(x) = g(a) + g'(a)(x - a)
where a is the given value (in this case, a = 2).
First, we need to calculate g(2) and g'(2) using the given information:
g(2) = 2f(2^2) = 2f(4) = 2*4 = 8
g'(2) = [f(2^2)*2] + [f'(2)*2(2^2)] = [f(4)*2] + [f'(2)*8] = (4*2) + (8*8) = 8 + 64 = 72
Therefore, with the given information, we have g(2) = 8 and g'(2) = 72.
Now we can substitute these values into the linearization formula:
L(x) = g(2) + g'(2)(x - 2)
L(x) = 8 + 72 (x - 2)
Simplifying further, we have:
L(x) = 8 + 72x - 144
Finally, the linearization L(x) is:
L(x) = 72x - 136
To find the linearization L(x) of the function g(x) = xf(x^2) at x = 2, we can use the formula for linear approximation:
L(x) = f(a) + f'(a) * (x - a)
where a is the point at which we want to find the linearization (in this case, a = 2).
First, let's find the values of f(2), f'(2), f(4), and f'(f):
f(2) = 1
f'(2) = 8
f(4) = 4
f'(f) = -3
Now, let's substitute these values into the linearization formula:
L(x) = f(2) + f'(2) * (x - 2)
Substituting the values we know:
L(x) = 1 + 8 * (x - 2)
Simplifying further:
L(x) = 1 + 8x - 16
Combining like terms:
L(x) = 8x - 15
So, the linearization L(x) of the function g(x) = xf(x^2) at x = 2 is given by L(x) = 8x - 15.