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Question

Find the value of f(0) if f(x + 2) = (x + 1)34 - (x + 1)33 + 5.

The correct answer is

7

The question asks us to find the value of the function at 0, which is f(0), given the equation relating f(x + 2) to x. The given equation is:

f(x + 2) = $(x + 1)^{34} - (x + 1)^{33} + 5$

To find f(0), we need the argument of the function f to be 0. In the given equation, the argument is $(x + 2)$. We need to find the value of $x$ that makes $(x + 2)$ equal to 0.

Set the argument equal to 0:

$x + 2 = 0$

Solving for $x$:

$x = -2$

Now that we have the value of $x$ that makes the argument of $f$ equal to 0, we substitute this value of $x$ (which is -2) into the right side of the given equation to find the value of f(0).

Substitute $x = -2$ into the expression $(x + 1)^{34} - (x + 1)^{33} + 5$:

f(0) = $(-2 + 1)^{34} - (-2 + 1)^{33} + 5$

Simplify the terms inside the parentheses:

f(0) = $(-1)^{34} - (-1)^{33} + 5$

Now, evaluate the powers of -1:

  • Any negative number raised to an even power is positive. So, $(-1)^{34} = 1$.
  • Any negative number raised to an odd power is negative. So, $(-1)^{33} = -1$.

Substitute these values back into the equation for f(0):

f(0) = $1 - (-1) + 5$

Simplify the expression:

f(0) = $1 + 1 + 5$

f(0) = $2 + 5$

f(0) = $7$

Thus, the value of f(0) is 7.

Function Evaluation Steps

Here are the steps we followed to find the value of f(0):

  1. Identify the expression given for f(x + 2).
  2. Determine the value of $x$ needed to make the argument (x + 2) equal to 0.
  3. Solve the equation $x + 2 = 0$ for $x$.
  4. Substitute this value of $x$ into the right-hand side of the given equation.
  5. Evaluate the resulting numerical expression.

Final Value Calculation

We found that for f(0), we need $x = -2$. Substituting $x = -2$ into the expression $(x + 1)^{34} - (x + 1)^{33} + 5$ gives:

$(-2 + 1)^{34} - (-2 + 1)^{33} + 5$

$= (-1)^{34} - (-1)^{33} + 5$

$= 1 - (-1) + 5$

$= 1 + 1 + 5$

$= 7$

So, f(0) = 7.

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Important Questions from Miscellaneous

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    2. Distance between the longitudes is maximum on the Equator.

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