if f(x)=x^2-1/x^3 then the slope of the tangent line at the point (1,0) is equal to;?

Question

if f(x)=x^2-1/x^3 then the slope of the tangent line at the point (1,0) is equal to;
a) 0
b) 5
c) -1
d) 1
e) non of these

Answer 1

the answer is b

Why - so the first thing to remember is that the derivative is the tangent line so all you have to do is find the derivative and then evaluate it when x=1

so what is the derivative
df/dx=2x+3/x^4

if you put x=1 into the above equation, df/dx=2+3=5=slope

Answer 2

I assume you mean:
f(x)=(x^2-1)/x^3
f'(x) = {x^3*2x - (x^2-1)3x^2}/ x^6 = (2x^2 -3x+3)/x^4
f'(1) = (2-3+3)/1 = 2, so answer is none of these.

If you truly meant x^2-1/x^3, then the derivative is f'(x) = 2x + 3/x^4, so f'(1) = 2 +3 = 5.

In the last case is the right 1 the answer is b)

Answer 3

B.

slope = 2X - (-3X^-4)

slope = 2 - (-3) = 5

Answer 4

f(x) =x^2 -x^-3
dx= 2x -3(-x^-4)=0
2(1)-3/(-1^4)=2-3/1=-1

the slope is -1

Answer 5

none of the above.
the slope is 6