Given the function y=3x^2+Bx-12, find the values for B such that the function is increasing when x=5 and decreasing when x=2?
I'm currently studying the limits and rates of change unit of grade 12 calculus, so I have to find the derivative only using the first principle formula.
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You need to remember that derivatives help you to study monotony of functions such that: if the first derivative is positive over an interval of values, then the function increases over the interval and if the derivative is negative over an interval, then the function decreases over the interval.
You suggest to find derivative using the first principl, hence you need to find derivative of function at x=5 such that:
`lim_(h-gt0) (3(x+h)^2 + B(x+h) - 12- 3x^2 - Bx + 12)/h`
You need to expand the binomial such that:
`lim_(h-gt0) (3x^2 + 6xh + 3h^2 + Bx + Bh - 12- 3x^2 - Bx + 12)/h`
Reducing like terms yields:
`lim_(h-gt0) (6xh + 3h^2+ Bh)/h`
Factoring out h yields:
`lim_(h-gt0) h*(6x + 3h+ B)/h`
Reducing by h yields:
`lim_(h-gt0) (6x + 3h+ B) = 6x + B`
`` Hence, evaluating the derivative of function using first principle yields f'(x) = 6x + B
You need to evaluate B if the function increases at x=5, such that:
`f'(5) = 6*5+B`
If the function increases at x = 5, then `f'(5) gt 0 =gt 30+Bgt0`
`B lt 30 =gt B in (-oo,30)`
You need to evaluate B if the function decreases at x=2, such that:
`f'(2) = 6*2+B`
If the function decreases at x = 2, then `f'(2)lt 0 =gt 12+Blt0=gtBgt12 =gt ` `B in (12,oo)`
Hence, evaluating the values of B when function increases at x=5 yields `B in (-oo,30` ) and when function decreases at x=2 yields `B in (12,oo). `
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