Question

Ja'Von kicks a soccer ball into the air. The function f(x) = –16(x – 2)2 + 64 represents the height of the ball, in feet, as a function of time, x, in seconds. What is the maximum height the ball reaches?

Answer 1

Answer:

64 feet

Step-by-step explanation:

Hi,

For a given function f(x) to have a maximum at any point P,

f'(x) should be equal to 0 and f''(x) < 0 at the point P.

Hence, given the height of the ball

f(x) = -16(x - 2)² + 64

f'(x) = 0

⇒ -32(x - 2) = 0

⇒ x = 2.

Consider f''(x) = -32 < 0

Hence, f(x) will have maximum at x = 2

Maximum height at x = 2 is -16(2 - 2)² + 64

= 64 feet

Hope, it helped !

Answer 2

Solution:

To find the maximum height of ball ,we must use application of derivative to find maxima.

here function of height represented as

$f(x) = - 16( {x - 2)}^{2} + 64 \\ \\ {f}^{'} (x) = - 16 \times 2(x - 2) + 0 \\ \\{f}^{'} (x) = - 32(x - 2)$

to find maxima

${f}^{'} (x) = 0 \\ \\ - 32(x - 2) = 0 \\ \\ - 32x + 64 = 0 \\ \\ 32x = 64 \\ \\ x = 2$
and for that

${f}^{''} (x)<0\\\\{f}^{''} (x)=-32$

So, maximum height is achieved when x = 2

put x= 2 in the f(x)

$f(2)= - 16( {2 - 2)}^{2} + 64 \\ \\ = 64 \: feet$
Hope it helps you.