Consider the function and the value of a.

Question

f(x) = 5 − x − x², a = 0

(a) Use M_tan= lim_x-0 (f(a+h) - f(a)) / (h) to find the slope of the tangent line m_tan = f'(a)

(b) Find the equation of the tangent line to f at x = a. (Let x be the independent variable and y be the dependent variable.)

Answer 1

For f(x) = 5 - x - x²:

1. f(a + h) = 5 - (a + h) - (a + h)² = 5 - a - h - a² - 2ah - h²
2. f(a) = 5 - a - a²
3. f(a + h) - f(a) = (5 - a - h - a² - 2ah - h²) - (5 - a - a²) = 5 - a - h - a² - 2ah - h² - 5 + a + a² = -h - 2ah - h²
4. [f(a + h) - f(a)]/h = (-h - 2ah - h²)/h = -1 - 2a - h
5. and the limit of this as h approaches zero is -1 - 2a

So m_tan (or f '(a) = -1 - 2a

The point-slope form of a line is y - y₁ = m(x - x₁) for the point (x₁, y₁). In this case, the point is (a, f(a)) but we have already determined that f(a) is 5 - a - a². Plugging in what we know then, into the point-slope form:

y - y₁ = m(x - x₁)

y - (5 - a - a²) = (-1 - 2a)(x - a)

You can simplify if you want to get it into whatever form you think is best. Perhaps y = (-1 -2a)x + a² + 5

Answer 2

f(x)=5-x-x²

a) I think there is a typo in the question. x→0 should be h→0 It is wanting to find f'(x) using the definition of a derivative first and then find m.

f'(x) = lim_h→0 (f(x+h)-f(x))/h

Using the 4 step process:

1) f(x+h) = 5-(x+h)-(x+h)² = 5-x-h-x^2-2xh-h²

2) f(x+h)-f(x) =5-x-h+x^2-2xh+h² - (5-x-x²) = -h-2xh+h²

3) (f(x+h)-f(x))/h = -1-2x+h

4) lim_h→0 (f(x+h)-f(x))/h = -1-2x

f'(x) = -1-2x

m=f'(a) = f'(0) = -1

b) Tangent line

f(0) = y₀ = 5

y-y₀ = m(x-x₀)

y = m(x-x₀)+y₀ = -1(x-0)+5 = -x+5