The graph of a function *f* is shown.

The *x* *y*-coordinate plane is given. The curve begins at *y* = 1 on the positive *y*-axis, appears to go horizontally right, passes through the approximate point (3, 1), goes up and right becoming more steep, and ends at the point (5, 3) nearly vertical.

Does *f* satisfy the hypotheses of the Mean Value Theorem on the interval [0, 5]?

Yes, because *f* is continuous on the open interval (0, 5) and differentiable on the closed interval [0, 5].Yes, because *f* is increasing on closed interval [0, 5]. Yes, because *f* is continuous on the closed interval [0, 5] and differentiable on the open interval (0, 5).No, because *f* is not differentiable on the open interval (0, 5).No, because *f* does not have a minimum nor a maximum on the closed interval [0, 5].No, because *f* is not continuous on the open interval (0, 5).

If so, find a value *c* that satisfies the conclusion of the Mean Value Theorem on that interval. (If an answer does not exist, enter DNE.)

c =

3

Community Answer

Step 1lGiven, The gnaph of a function 'f'Here f(x) is continuous on x in(0,5) But not on x=0, and 5 because x rarr0^(-)&quadx^(a)rarr5^(+)' f ' is not deterinined.check difterentiatility on x in(0,5)Here f(x) has two different type of graph (one is streight line 8 second one is stee-p curve). So, we have to check at joint point of woth curve. (Jomt point x=3 )f(x) is strenght une & panallel to x-axis on the left of x=3So f^(')(x rarr3^(-)) ... See the full answer