Question

8) (8 pts total) For this problem, you will sketch a graph of f(x) = 2x4 + 8x3. Complete the following steps: (a) (1 pt) Determine the intercepts of the function. (b) (3 pts) Use the first derivative to find the intervals on which f increases and decreases, and the relative maximums and minimums. (c) (3 pts) Use the second derivative to find the intervals on which f is concave up and concave down, and the inflection points. (d) (1 pt) Bring all of this information together to sketch a graph of this function. Use the provided axes at the bottom of this page, or use the same scales if drawing your own.

Answer 1

given that f(a) a 2x4 +843 (1) lie. We have to find intercepts of the function for y intercept putting x=0 y = f(a) = 0 & for X-intercepts putting y=falco - 2x4 +8x3 -0 3) 273(x+4)=0 > Thus, The The function, fon) cuts x-axis at (60) 8 (4,0) Cooms y-axis at (0,0) 220, x= -4 (Ь f(x) = 8x3 + 24² let f'(as so »):8.834247270 3) 822(x+3) > 0 ht, - inta o her fearo) i's increasing in (-00,-3) U1 0.00) is decreasing in (-3,0) falar) at X=0 of min 2472 +48 X f (o)= 0 No conclusion of local max f''(x) 482 +48 f"(0) - 48 + 0 ) (0) is a point of inflection at a = - 4 f" (-4) 24x.16 + 481-4) 48 x 8«t 48 (-4) > 0 a : :-4 is a local maxina minima &

let f'(x) > 0 24 22 +48 xso » 24% (7+2) > 0 aty - tete ne fox) is Concare up in (-00,-2) Cores) & Concave down (-2,0) y -4 -3 -27 2 t