Question

1 point
Find the point c in the curve f(x) = x^3 + x^2 + x + 1 in the interval [0, 1]
where slope of a tangent to a curve is equals to the slope of a line joining
(0.1) *
O 0.64
O 0.54
O 0.34
M
O 0.44
Find the value of c which satisfies the Mean Value Theorem for the given
17 point​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

The point c in the curve f(x) = x³ + x² + x + 1 in the interval [0, 1]

where slope of a tangent to a curve is equals to the slope of a line joining. (0, 1)

• 0.64

• 0.54

• 0.34

• 0.44

Find the value of c which satisfies the Mean Value Theorem for the given points

EVALUATION

Here the given function is f(x) = x³ + x² + x + 1 in the interval [0, 1]

Now

(i) f(x) is continuous in the interval [0,1]

(ii) f(x) is differentiable in (0,1)

So Mean Value Theorem is applicable for the given function f(x) = x³ + x² + x + 1 in the interval [0, 1]

By the mean value theorem there exists at least one value of x say c in (0,1) such that

$\displaystyle \sf{f'(c) = \frac{f(1) - f(0)}{1 - 0} }$

Now f'(x) = 3x² + 2x + 1

f'(c) = 3c² + 2c + 1

f(1) = 1³ + 1² + 1 + 1 = 4

f(0) = 0³ + 0² + 0 + 1 = 1

Thus we get

$\displaystyle \sf{f'(c) = \frac{f(1) - f(0)}{1 - 0} }$

$\displaystyle \sf{ \implies \: 3 {c}^{2} + 2c + 1 = 4 - 1}$

$\displaystyle \sf{ \implies \: 3 {c}^{2} + 2c - 2 = 0}$

$\displaystyle \sf{ \implies \: c = \frac{ - 2 \pm \: \sqrt{4 - 4 \times 3 \times ( - 2)} }{2 \times 3} }$

$\displaystyle \sf{ \implies \: c = \frac{ - 2 \pm \: \sqrt{4 + 24} }{2 \times 3} }$

$\displaystyle \sf{ \implies \: c = \frac{ - 2 \pm \: \sqrt{28} }{6} }$

$\displaystyle \sf{ \implies \: c = \frac{ - 2 \pm \: 5.29}{6} }$

⇒ c = 0.54 , - 1.2

Since 0 < c < 1

So c = 0.54

Hence the required value of c = 0.54

FINAL ANSWER

Hence the correct option is 0.54

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