$f'(x)\ge0$ for all $x$ in [-3, 0], so $f(x)$ is non-decreasing over this interval, and in particular we know right away that its minimum value must occur at $x=-3$ .

From the plot, it's clear that on [-3, 0] we have $f'(x)=-x$ . So

$f(x)=\displaystyle\int(-x)\,\mathrm dx=-\dfrac{x^2}2+C$

for some constant $C$ . Given that $f(0)=7$ , we find that

$7=-\dfrac{0^2}2+C\implies C=7$

so that on [-3, 0] we have

$f(x)=-\dfrac{x^2}2+7$

and

$f(-3)=\dfrac52=2.5$

5 0

3 years ago

Graphing points for y=|2x-1|<br> Pls help first person to answer get the Brainly points

Olin [163]

Answer:

X-intercept= (1/2,0) Y-intercept= (0,1)

Step-by-step explanation:

You're welcome :)

8 0

3 years ago

This is for my friend so

Alex

What do you need help wiht ?
It’s practically d

5 0

2 years ago