Math Master's Please Help me...?

Question

Consider the family function f(x) = e^x + k. What value of k gives the curve that is tangent to the line y = 4x+5?

I have tried solving this by taking the derivative of f(x) and sets it equals to 0 but it didn't work because k becomes 0 if I take it to 0, what should I do??

Answer 1

That's interesting, I was JUST reading up about Lagrange multipliers yesterday. Here's the wikipedia link on "how to", the instructions are pretty clear-cut. The example they use is very simliar to your problem.

Addendum: The other guys here have the right approach for functions of one variable.

Answer 2

You need to find the point at which f(x) touches 4x + 5.

For the line 4x + 5 to be tangent to f(x) it must touch at just one point and have the same slope.

You can differentiate y = 4x + 5 to see that the line has slope 4. So where does f(x) have slope 4? When f'(x) = 4. (Your f'(x) = 0 is a red herring.)

Now at the point where f'(x) = 4, you just need to make sure that f(x) = 4x + 5 - which will give you k.

Answer 3

Ah ok simple.
First you need to find at what X value f(x) has a slope of 4. So differentiate f(x) and you get:
f'(x)=e^x
so f'(x)=4 when x = ln(4)
now you use that to find at what value on your tangent the y is at this X value:

y=4ln(4)+5

now you substitue that in for f(x) of your function and ln(4) for X and solve for k

4ln(4)+5= e^ln(4)+k
4ln(4)+5= 4 + k
4ln(4)+1= k

Answer 4

Tee hee.

e^x is the only function whose slope equals it's value!,

So, you want to adjust k so that

e^x+k = 4x+5
and
e^x = 4, the slope of the line.