Question

Find the maximum y-value on the graph of y=f(x)?

f(x)=-x^2+2x+1

this is what I got.
f (x) = - 2*x + 2
- 2*x + 2 = 0
x = 1
f(1) = - 1 + 2*2 + 1 = 2, Max

Answer 1

your first derivative is in-correct.

Answer 2

\[
f(x)=x^2+2x+1\\
f'(x)=2x+2=0\implies x=-1\rightarrow\text{critical point}\\
f''(x)=2>0\rightarrow\text{condition is for minima}
\]
the function has a local minima and no relative maxima

Answer 3

I dont get it. sigh

Answer 4

we have
\[f(x)=x^2+2x+1\]
what is its first derivative?

Answer 5

@marcybaby the first derivative will be
\[f'(x)=2x+2\]

Answer 6

- 2*x + 2 = 0?

Answer 7

why the -2x?

Answer 8

ooooh!!!!! I see messing up.

Answer 9

how did you get the negative sign?

Answer 10

I guess I keep looking that the problem.

Answer 11

the answer to the problem is that there is no relative maxima for the function

Answer 12

how come? thats the part I am having issues understanding.

Answer 13

@marcybaby so you see,
\[f'(x)=2x+2\]
to find the critical points, we set this equal to zero
\[2x+2=0\\
2x=-2\\
\boxed{x=-1}\]

Answer 14

so, we have one critical point x=-1
we do not know yet if this is a point of maxima or minima

Answer 15

to check what it is, we take the help of second derivative
\[f'(x)=2x+2\\
\text{differentiate with respect to x}\\
f''(x)=2\\
\text{plug in the critical value in this equation}\\
f''(-1)=2
\]
the property tells us that if:
1) f''(x) > 0 -----> means f(x) is minimum at that critical point
2) f''(x) < 0 ------> means f(x) is maximum at that crititcal point
3) f''(x) = 0 ------> means that the test failed to verify and f(x) is stationary at that critical point

Answer 16

i see!