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# Calculus help! Please help me.. Please!! Find the equation of the tangent line of the given function at x=1.?

1) f(t)=t^3 -6t=1 ------------ I got y=-2x-2

2) g(x)= 2x^5 + square root of x ------ I got y=39.5x-36.5

3)h(y)=x + (2/x) - (3/x^2) --------- I got y=5x-5

4) f(x)=x^8 - cube root of x + 6x^4

5) p(t)=2t^5 - 5t^2 + (1/square root of x) -8 -------y=-3x-7

Are my answers correct? If they aren't, please explain why and the steps to get the correct answer! Thank you so much.

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- Anonymous8 years agoFavorite Answer
A problem at a time...

f(t) = t³ - 6t + 1

f'(t) = 3t² - 6

If t = 1...

f(1) = -4

f'(1) = -3, so?

y + 4 = -3(x - 1)

y + 4 = -3x + 3

y = -3x - 1

Redo that problem before going to the next problem. You need more practice with differentiation!

Source(s): τ

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