Answer:
L(x)=-4x+9
L(2.1)=0.6
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.