local linerarization is equal to la (x)=2x-2 what is the value of a? what is the value of f(a)? Use the linearization to approximate the value of f(5.6)?

Question

link to graph: https://webwork.utrgv.edu/wwtmp/Calculus_I_Pena/images/3dffd8ca-2b45-3f51-9432-1079aa714a3b___b2c3bc82-9722-3765-ae1c-e6a6da7ac10c.png

Answer 1

The problem and graph do not fit together for this problem.

We know both from graph and f'(x)=2x-2 that f(x) is a quadratic.

from the graph information we have f(x)=x(x-5)+C,,,this is the problem

now f'(x)=2x-5

let's say C=15 then f(a)=f(5)=15

,,,then the approximation for f(x) for 5.6 would be approx=15+[2(5)-5](5.6-5)=15+5(.6)=18

the actual f(5.6)=5.6(.6)+15=18.36

Answer 2

As shown from this source, https://tutorial.math.lamar.edu/classes/calci/linearapproximations.aspx, the formula for linear approximation is

L(x) = f(a) + f'(a)(x-a).

The graph shows that the approximation and function intersect at x=5, since by definition of our tangent / linear approximation line, they intersect at (a, f(a)) [see the link above]. That 'a' value is the x coordinate for the intersection, which is 5 :)

a = 5

We also know that f'(a) = f'(5) = 2 because that is the slope aka 'm-value' of our line from the L(x) = 2x-2 equation. Think of y-intercept form y = mx + b or point intercept form y = y₁ + m(x-x₁), either way, the coefficient is 'm'.

If we plug back in a and f'(a) we can solve for f(a).

L(x) = f(a) + f'(a)(x-a)

2x - 2 = f(a) + f'(a)(x-a)

2(5) - 2 = f(a) + 2 (5-5)

10-2 = f(a) + 0

f(a) = 8

Now we can either use either simplification of L(x) to find the approximation.

L(x) = f(a) + f'(a)(x-a) = 8 + 2(x-5) or

L(x) = 2x-2

I'd use the latter for simplicity.

f(5.6) ≈ L(5.6) = 2(5.6) - 2 = 11.2 - 2 = 9.2

Hope this helps!