Find the minimum and maximum values of y=6θ−3sec⁡θ on the interval [0,π3]

y=6θ−6sec⁡θ on the interval [0,π/3])f(0)=−6f(π3)=6(π3)−3sec⁡(π3)=6(π3)−3(2)=−0.899f′(x)=6−3sec⁡(θ)tan⁡(θ)=06=3sec⁡(θ)tan⁡(θ)sec⁡(θ)tan⁡(θ)=63θ=π4f(π4)=6(π4)−3sec⁡(π4)=−0.52566we have: f(0)=−6=−2.4495f(π3)=−0.899f(π4)=−0.52566minimum =−(6)(occurs at x=0)maximum =−0.52566(occurs at x=π/4)

Answered: "Find the minimum and maximum values of y=6θ−3sec⁡θ..."

High School
Algebra
Pre-Algebra
Negative numbers and coordinate plane

Globokim8

Answered question

2021-08-11

Find the minimum and maximum values of $y = \sqrt{6} θ - \sqrt{3} \sec θ$
on the interval $[0, \frac{π}{3}]$

Answer & Explanation

Nichole Watt

Skilled2021-08-12Added 100 answers

$y = \sqrt{6} θ - \sqrt{6} \sec θ on the interval [0, π / 3])$
$f (0) = - \sqrt{6}$
$f (\frac{π}{3}) = \sqrt{6} (\frac{π}{3}) - \sqrt{3} \sec (\frac{π}{3}) = \sqrt{6} (\frac{π}{3}) - \sqrt{3} (2) = - 0.899$
$f^{'} (x) = \sqrt{6} - \sqrt{3} \sec (θ) \tan (θ) = 0$

$\sqrt{6} = \sqrt{3} \sec (θ) \tan (θ)$
$\sec (θ) \tan (θ) = \frac{\sqrt{6}}{\sqrt{3}}$
$θ = \frac{π}{4}$
$f (\frac{π}{4}) = \sqrt{6} (\frac{π}{4}) - \sqrt{3} \sec (\frac{π}{4}) = - 0.52566$
we have: $f (0) = - \sqrt{6} = - 2.4495$
$f (\frac{π}{3}) = - 0.899$
$f (\frac{π}{4}) = - 0.52566$
minimum = $- \sqrt{(} 6) (occurs at x=0)$
maximum = $- 0.52566 (occurs at x = π / 4)$

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