Please help! We know that f(2) = 3, f′(2) = 5. Approximate f(2.15) using the tangent line approximation generated by the given information. Is f an increasing function near x = 2?

Question

anonymous · Answer

$$f(x)\approx f'(x)(x-x_0)+f(x_0)$$

jim_thompson5910 · Answer

In general, the equation of the line tangent to f(x) at the point x = a is

y = f(a) + f ' (a) * (x - a)

jim_thompson5910 · Answer

In your case, a = 2, so the tangent line is

y = f(2) + f ' (2) * (x - 2)

anonymous · Answer

$$x_0=2,f(x_0)=f(2)=3$$

anonymous · Answer

Jim, thank you. But what does it mean to approximate f(2.15)?

jim_thompson5910 · Answer

once you have the tangent line, you can use it to approximate f(2.15)

this is because 2.15 is relatively close to 2

anonymous · Answer

Thank you. I'm using a program called GeoGebra. I'll try to take your suggestions and draw this out - If possible, may I contact you within the next 10 minutes or so if I have additional questions?

jim_thompson5910 · Answer

sure, but no need to use geogebra since you really don't have a function to work with

just use the numbers given

y = f(2) + f ' (2) * (x - 2)

y = 3 + 5(x - 2)

y = 3 + 5x - 10

y = 5x - 7

So the equation of the tangent line to f(x) at x = 2 is y = 5x-7

So at and around x = 2, the function f(x) is approximately g(x) = 5x-7

So to find f(2.15), evaluate g(2.15)

g(x) = 5x-7

g(2.15) = 5(2.15)-7

g(2.15) = 10.75-7

g(2.15) = 3.75

Therefore, f(2.15) is approximately 3.75

anonymous · Answer

My goodness, I appreciate you so much! I need to stare at this for a few moments to fully understand what I am seeing, but it's slowing starting to make sense. THANK YOU!

jim_thompson5910 · Answer

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