Find the local linear approximation L(x) of the function f(x) = 5−x^2 at x = 2. Use this to estimate f(2.1).

L(x)=-4x+9

L(2.1)=0.6

Step-by-step explanation:

It's asking us to find the tangent line to curve f(x) = 5−x^2 at x = 2.

Theb use this to estimate f(2.1).

To find slope of tangent line, we must differentiate and then plug in 2 for x.

f'(x)=0-2x by constant and power rule.

f'(x)=-2x

So the slope of the tangent line is -2(2)=-4.

A point on this tangent line shared by the curve is at x=2. We can find it's corresponding y-value using f(x)=5-x^2.

f(2)=5-(2)^2

f(2)=5-4

f(2)=1

So let's rephrase the question a little.

What's the equation for a line with slope -4 and goes through point (2,1).

Point-slope form y-y1=m(x-x1) where m is slope and (x1,y1) is a point on the line.

Plug in our information: y-1=-4(x-2).

Distribute: y-1=-4x+8

Add 1 on both sides: y=-4x+9

Let's call this equation L(x), an expression to approximate value for f near x=2.

L(x)=-4x+9

Now the appropriation at x=2.1:

L(2.1)=-4(2.1)+9

L(2.1)=-8.4+9

L(2.1)=0.6

If we did plug in 2.1 into given function we get 5-(2.1)^2=0.59 . This is pretty close to our approximation above.

why can't you?

step-by-step explanation: