Can someone answer calculus question?

1.

Let z=xy

y = 2x-8

z = x(2x-8)

z = 2x² - 8x

dz/dx = 4x - 8

dz/dx = 0 when x=2

d²z/dx² = 4 < 0

This is a minimum

When x=2, y = -4

xy = -8

The minimum value of xy is -8

B

2.

f(x) will have a maximum/minimum value when f'(x)=0

f'(x) = e^x - 3x²

f'(-0.46) ≈ 0.949

f'(-0.20) ≈ 1.101

f'(0.91) ≈ 0

f'(0.95) ≈ -0.121

f'(3.73) ≈ -0.060

f(x) will have a relative maximum/minimum value at 0.91

Check second derviative:

f"(0.91) ≈ -2.97 < 0

It is a local maximum

C

I agree with mastermind about #1, but not entirely about #2.

For #2, if f'(x) = e^x-3x62, then f(x) = e^x - x^3. One way to answer the question is to graph the function. If you do, you can see that there is a local max (a place where it looks like the top of a mountain) at x = ..91.

1)

y=2x -8

xy =x(2x-8)

2x²-8x

dy/dx=4x-8

0=4x-8

x=2

y=2(2)-8

y=-4

xy=2(-4)= -8

Answer is B. -8

2.max value f''(x) < 0

f"(x) = e^x - 6x

e^x -6x <0

e^x < 6x

when x = 0.95 , 0.91

e^x < 6x

Answer is C and D

Failed algebra 2 so can't help u