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Derivative of Standard Functions

If fx = x - 1...

Question

If f(x) = x - 1, g(x) = 2x + 3, and h(x) = 5x + 4, then what is the value of fogoh(1)?
20

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Solution

The correct option is A 20
f(x) = x - 1, g(x) = 2x + 3, and h(x) = 5x + 4
fogoh(x) = f(goh(x)) ... (i)
goh(x) = g(h(x))
goh(x) = g(5x + 4)
goh(x) = 2(5x + 4) + 3
goh(x) = 10x + 11
put goh(x) in equation (i)
fogoh(x) = f(10x + 11)
fogoh(x) = 10x + 11 - 1
fogoh(x) = 10x + 10
So, fogoh(1) = 10 + 10 = 20

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Similar questions

f (x) = \frac{2}{x - 3}

g (x) = \frac{x - 3}{x + 4}

h (x) = - \frac{2 (2 x + 1)}{x^{2} + x - 12}

lim x \to 3 [f (x) + g (x) + h (x)]

f (x) = \frac{2}{x - 3}, g (x) = \frac{x - 3}{x + 4}

h (x) = \frac{- 2 (2 x + 1)}{x^{2} + x - 12}

lim x \to 3 [f (x) + g (x) + h (x)]

f (x) = 2 x + | x |, g (x) = \frac{1}{3} (2 x - | x |)

h (x) = f (g (x))

h (1)

f (x) = x^{2} + \frac{1}{x^{2}}

g (x) = x - \frac{1}{x}, x ϵ R - {- 1, 0, 1}

h (x) = \frac{f (x)}{g (x)}

. Then the local minimum value of

h (x)

f (x) = x^{2} - \frac{1}{x^{2}}

g (x) = x - \frac{1}{x}, x \in R - - 1, 0, 1

h (x) = \frac{f (x)}{g (x)}

then the local minimum value of

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