Applications of derivatives

Question

The tangent to f(x)=(2x-1)^4 at x=k has gradient 8. 

 

a) find k

b) find equation of tangent 

c)hence find the x intercept of this tangent

Answer 1

f(x) = (2x-1)⁴

 

f'(x) = 4(2x-1)³(2)

 

Since f'(k) = 8, we have 8(2k-1)³ = 8.   So, 2k-1 = 1

 

                                                                    k = 0

 

Point on tangent line:  (k, f(k)) = (0, 1)

 

Slope of tangent line:  f'(k) = 8

 

Equation of tangent line:  y = 8x + 1

 

x-intercept:  set y = 0 and get x = -1/8       x-int = (-1/8, 0)