carleyschlag carleyschlag
  • 11-10-2021
  • Mathematics
contestada

Show that f(x)=12x^3-32x^2+17 has a zero between 2 and 3.

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jcherry99
jcherry99 jcherry99
  • 11-10-2021

Answer:

Step-by-step explanation:

You could graph it. I'll do that at the bottom. What you need is a value that is - on either 2 or 3 and a plus on the other one.

y=12*2^3 - 32x^2 + 17

y = 12 * 8 - 32*2^2 + 17

y = 96 - 128 + 17

y = -15

Now when x = 3 should give you something above the x axis

y = 12*3^3 - 32*9 + 17

y = 324 - 288 + 17

y = 53

What this indicates is that somewhere between 2 and 3 the graph crosses the x axis. The graph should indication that.

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