Answer the following questions for the function f ( x ) = x √ x 2 + 25 defined on the interval − 4 ≤ x ≤ 7 .

Question

Answer the following questions for the functiondefined on the interval . is concave down on the interval x =    to x =    is concave up on the interval x =    to x =   The inflection point for this function is at x =   The minimum for this function occurs at x =   The maximum for this function occurs at x =

Answer 1

f'(x) = √x² + 25 + x²/√x² + 25 = (2x² + 25)/√x² + 25 > 0 for all x. So, f(x) is increasing function on interval [-4, 7].

So min f(x) = f(-4) = - 4√41 and max f(x) = f(7) = 7√74;

f''(x) = (4x√x² + 25 - (2x² + 25)·x/√x² + 25)/(x² + 25) = (4x³ + 100x - 2x³ - 25x)/(x² + 25)^3/2 =

(2x³ + 75x)/(x² + 25)^3/2 = 0; 2x³ + 75x = 0; x = 0.

Now for x⊂ (-4, 0) f''(x) < 0 and graph concave down; for x⊂ (0, 7) f''(x)> 0 and graph concave up.

f(0) = 0 and (0, 0) is inflection point