Tangent Line Calculus

Question

I have two questions that I'm stuck on and they both involve finding the tangent line.

For the first one, I already got part a) right but I'm struggling with part b).

Compute 𝑓′(𝑥) and find an equation of the tangent line to the graph at 𝑥 = 𝑎. 

𝑓(𝑥)=3𝑥−32sqrt(x) a=4

(𝑎) 𝑓′(𝑥)= 3-(16)/((x^(1/2)))

(𝑏) Tangent line to the graph has equation 𝑦= 

The second problem

I already got part a) but I need help on part b).

Let 𝑓(𝑥)=1(3sqrt(x)) (the equation is 1 times the cube root of x)

Compute 𝑓′(27) and find an equation of the tangent line to the graph at 

𝑥 = 27.

(Use symbolic notation and fractions where needed.)

a) 𝑓′(27) = 1/(3x^(2/3))

b) The equation of the tangent line at 𝑥=27 is 𝑦 = 

Thanks!

Answer 1

1. The easiest way to find the tangent line is to use point-slope formula from Algebra 1.

y - y₁ = m ( x - x₁ )

You are told to use the point on the graph x=4. Well every point needs both an x-value and a y-value so you need to plug x=4 into the equation 𝑓(𝑥)= 3𝑥−32√x to get its corresponding y-value. Use that point (4, ___) as your (x₁, y₁) in the point-slope formula above.

Now you have your point, so to finish point-slope form the only thing you need is the slope (m). Remember that the derivative IS the slope at any point. You already found the formula for the derivative, but we specifically want the derivative when x=4. So use your formula 𝑓′(𝑥)= 3-(16)/((x^(1/2))) and plug in x=4. Whatever comes out (f'(4)= ___) is your m-value.

Now you have y - ____ = ____ ( x - ____ )

^y₁ ^m ^ x₁

If you want the form y= you just need to do some algebraic rearranging (distribute m and then add y₁ to both sides). The process is exactly the same for your second question. Hope this helps!

Answer 2

The tangent to the graph f'(a)=[y-f(a)]/(x-a).

The answer is the same to both parts, just with different f's and different a's.