stumped: Let f be fcn on interval [-3, 7] with f(2) = 3.

[mickeymouse]

any possible help/ hints???

i always have difficulty answering questions referring to the graph of f(x) when the problem gives ONLY the graph of f'(x)
see below picture. click for enlarged version

:?:

 

[stapel]

For those who cannot see the image, the exercise is as follows:

\(\displaystyle \text{Let }f \text{ be a function defined on the closed interval }[-3, 7]\)
\(\displaystyle \text{ with }f(2)\, =\, 3.\)

\(\displaystyle \text{The graph of }f' \text{ consists of a line segment from }(-3, 0)\)
\(\displaystyle \text{ to }(-1, -2)\text{, from }(-1, 2) \text{ to }(2, 1) \text{, and from }(6, 1) \text{ to }(7, 0),\)
\(\displaystyle \text{ with the second and third segments "joined" by the upper}\)
\(\displaystyle \text{half-circle }(x\, -\, 4)^2\, + (y\, -\, 1)^2\, = 4 \text{, as shown in the graphic.}\)

\(\displaystyle \text{a) Find }f(-3)\text{ and } f(7).\)

\(\displaystyle \text{b) Find an equation for the line tangent to the graph at }(2, 3)\).

\(\displaystyle \text{c) On what interval is }f\text{ increasing? Justify your answer.}\)

\(\displaystyle \text{d) On what interval is }f\text{ concave up? Justify your answer.}\)

 

[skeeter]

hints for (a) ...

\(\displaystyle f(2) - f(-3) = \int_{-3}^2 f'(x) \, dx\)

\(\displaystyle f(7) - f(2) = \int_2^7 f'(x) \, dx\)

(b) f'(2) is the slope of the tangent line ... (2,3) is the point ... remember the point-slope form?

(c) f is increasing wherever f'(x) > 0, right? on what intervals is f'(x) positive?

(d) f is concave up whenever f''(x) > 0 ... note that f''(x) is the slope of f'(x) ... on what intervals is the slope of f'(x) positive?