Question

find the value of c if f(x)=x(x-3) e^3x,is continuous over interval [0,3] and differentiable over interval (0,3) and C €(0,3).​

Answer 1

We have to find the value of c , if f(x) = x(x - 3)e⁽³ˣ⁾, is continuous over interval [0, 3] and differentiable over interval (0, 3) and c ∈ (0, 3)

The conditions of Mean value theorem are,

• f(x) must be continuous on [a, b]
• f(x) must be differentiable on (a, b)

then there must be a point c on (a, b) such that,

$f'(c)=\frac{f(b)-f(a)}{b-a}$

here, a = 0 and b = 3

f(0) = 0(0-3)e⁽⁰⁾ = 0

f(3) = 3(3 - 3) e⁹ = 0

now differentiate f(x) with respect to x, we get

f'(x) = (2x - 3)e⁽³ˣ⁾ + (x² - 3x)3e⁽³ˣ⁾

$\implies f'(c)=(2c-3)e^{3c}+(c^2-3c)3e^{3c}$

from the mean value theorem,

$f'(c)=\frac{f(3)-f(0)}{3-0}=\frac{0-0}{3-0}=0\\\implies f'(c)=(2c-3)e^{3c}+(c^2-3c)3e^{3c}=0\\\implies(3c^2-9c+2c-3)e^{3c}=0\\\implies 3c^2-7c-3=0\\\implies c\approx-0.37,2.7$

but c ∈ (0,3) so, c ≠ - 0.37 hence, c = 2.7

Therefore the value of c is 2.7