1 of 8 0-day streak

Read and construct a gradient function from a function graph

Question 1 of 8

No calculator

The function and its relevant graph features are given in exact form

f(x)=\left(x-1\right)\left(x-5\right)^2

The x-intercepts are \(x=1,\quad 5\)

The relevant stationary or corner x-values are \(x=\frac{7}{3},\quad 5\), with points \((\frac{7}{3},\frac{256}{27}),\quad (5,0)\)

From left to right, the function has positive slope on x < 7/3; negative slope on 7/3 < x < 5; positive slope on 5 < x

(a)1 mark

Select the interval sign information that follows from the exact graph features

The gradient is zero when \(x=\frac{7}{3},\quad 5\)

f'(x)>0\text{ for }x<\frac{7}{3}\quad f'(x)<0\text{ for }\frac{7}{3}<x<5\quad f'(x)>0\text{ for }5<x

The gradient has degree 3

The gradient is zero when \(x=1,\quad 5\)

f'(x)>0\text{ for }x<1\quad f'(x)<0\text{ for }1<x<5\quad f'(x)>0\text{ for }5<x

The gradient is zero when \(x=\frac{7}{3},\quad 5\)

f'(x)<0\text{ for }x<\frac{7}{3}\quad f'(x)>0\text{ for }\frac{7}{3}<x<5\quad f'(x)<0\text{ for }5<x

The gradient is zero when \(x=\frac{7}{3},\quad 5\)

f'(x)>0\text{ for }x<\frac{7}{3}\quad f'(x)<0\text{ for }\frac{7}{3}<x<5\quad f'(x)>0\text{ for }5<x

(b)1 mark

Enter the x-coordinate of the first listed turning or corner point

x-coordinate of first feature point